USACO 终于做完了

用了大概一个半月的时间吧，当然中间有大概一个礼拜空着来着

这是第二次做USACO了，第一次是高中的时候，那个时候的题目跟现在的都不太一样，主要是顺序，而且那个时候是用PASCAL写的

但是高中的时候没有做完，卡在了Section 5之前，其实是因为很多东西不会，数学其实也不够好，至于理解的能力，不知道现在是不是也有所提高了

其实这次做的并不是非常顺利，我不是牛人，不可能一天扫10几道题目那种，然后每到题目半个小时就搞定

前面3个Section的题目还都不算是很难，都是训练型的，都是教你算法怎么用，从Section 4开始就有比较难的题目了，尤其是DP

DP从高中开始我就没有感觉，那时候就有人跟我说，只能多做题目，也许我现在做的还不够多吧，总感觉DP是很有用很有用的东西，所以一直想学好

不管怎么样，USACO总算是磕磕碰碰的做完了，应该在NoCow和Google的帮助下终于做完了

后面大部分题目我都看了解题报告，有些算法想得出，但是不知道该怎么应用到题目之上

现在发现这点才是最重要的，算法模块谁不会写啊，都可以提前写好一个放在那里，但是问题是怎么把这个算法应用到题目上

可能这就是所谓的建模吧，把题目变形一下，然后跟我们已知的算法联系起来

还有另外一种情况，这个题目的算法不是已知的任何算法，要自己去想的，这才是真正考验一个人算法素养的时候

就像TCO里面的题目，其实很多都是这样的，很少会给你一道题目让你直接去套一个算法的

可能原来我在这方面的理解就有偏差，我总认为，你把所有的常见算法都练熟了，所有的题目都可以横扫

但是问题就是，你能不能看得出来哪道题目用哪种算法

而且，就像DP这种题目，就算你看出来了，状态转移方程你也未必写的出来

总之还是学到了很多，虽然磕磕碰碰，但是做完了100道题还是会有收获的

不知道下一个目标是什么SGU呢，还是TCO

其实之所以喜欢USACO的一个原因就是他会告诉你测试数据，你可以很方便的Debug

像OJ这种，不告诉你测试数据的，如果遇到了WA，你就要想破脑袋去想你的程序哪里错了

而往往一个人自己看自己的程序的时候，是很难发现错误的，这就会让人很郁闷，真的是非常郁闷

更何况，有些OJ真的是很贱的，用一些超级恶心的数据来钻你的空子。

不知道这样对不对，也许是考验你的细心程度吧

SGU or TCO 呢？

posted @ 2009-05-11 09:27 杨磊阅读(1220) | 评论 (1) | 编辑收藏

USACO Section 5回顾

TEXT Convex Hulls	突包
PROB Fencing the Cows	突包问题，直接忽略，差不多所有的计算几何我都跳过了
PROB Starry Night	超级麻烦题，用Flood Fill把所有的pattern全认出来，然后判断重复，不重复就添加新的
PROB Musical Themes	DP题，技巧在于如果两个序列是theme，那么相邻的两个number差相等
PROB Snail Trail	DFS，没啥技巧
PROB Electric Fences	刚开始以为是Divide&Conquer，后来发现跟那道题不一样，直接搜索就可以
PROB Wisconsin Squares	搜索题，Testcase只有sample那一组
TEXT Heuristics & Approximate Searches	启发式搜索，没看
PROB Milk Measuring	一道我看Analysis看了两天的DP，其实还是没有深刻理解
PROB Window Area	可以用矩形切割过
PROB Network of Schools	强连通分量题
PROB Big Barn	最大子正方形，简单的DP
PROB All Latin Squares	搜索+剪枝
PROB Canada Tour	诡异的DP算法，Analysis的那个DP倒是还可以接受，不过Nocow上的似乎需要证明，但是又没有
PROB Character Recognition	这道题目都能用DP，不得不感叹DP的伟大
PROB Betsy's Tour	又是搜索+剪枝
PROB TeleCowmunication	先把点转化成边，然后求最小割
PROB Picture	离散化
PROB Hidden Passwords	搜索+KMP优化
PROB Two Five	算是道难题吧，Analysis都看了好久，DP真是无所不能

第六个Section的就不写了，两个DP+一个牛叉的位运算

那个Cow XOR我估计我这辈子都不会忘记了。

posted @ 2009-05-11 09:16 杨磊阅读(572) | 评论 (1) | 编辑收藏

USACO Section 4回顾

TEXT Optimization Techniques	讲怎么剪枝的
PROB Beef McNuggets	初看上去像是一道背包问题，但是用背包肯定超时，后来看了解题报告，发现原来是数学题
PROB Fence Rails	高维背包问题，只能搜索
PROB Fence Loops	其实是很简单的一道最短路问题，恶心就恶心在图的转化
PROB Cryptcowgraphy	非常恶心的搜索+剪枝
TEXT "Network Flow" Algorithms	网络流，我第一次会写网络流就是看了这个算法
PROB Drainage Ditches	网络流练习题
PROB The Perfect Stall	最大匹配，匈牙利算法
PROB Job Processing	第一问是贪心，第二问应该也还是贪心，就是把第一问最快做完的给第二问最慢做完的
PROB Cowcycles	直接枚举的好像
TEXT Big Numbers	高精度
PROB Buy Low, Buy Lower	经典DP，最长下降序列，可是问题是要求出现了多少次，于是我看了解题报告
PROB The Primes	搜索+剪枝，要注意搜索的顺序，先是第五行第五列，然后对角线，然后其他
PROB Street Race	关键路径，去掉每一个节点，然后看看起点与终点是否连通，不联通总说明是关键节点
PROB Letter Game	枚举，分两块，先找完整的单词，然后找pair
PROB Shuttle Puzzle	刚开始以为搜索，后来看了解题报告，发现原来有规律的，寒啊
PROB Pollutant Control	最小割问题
PROB Frame Up	搜索题，用一张表来维护每个pattern的上下关系，可以大量剪枝

posted @ 2009-05-11 08:52 杨磊阅读(355) | 评论 (0) | 编辑收藏

USACO Section3回顾

TEXT Minimal Spanning Trees	最小生成树，经典的算法
PROB Agri-Net	最小生成树，USACO这点比较好，一般讲完了一个算法，都会出一道练习题
PROB Score Inflation	背包问题
PROB Humble Numbers	经典题目，算法是用已有的丑数乘上集合里面的素数去生成新的丑数
PROB Shaping Regions	记得高中的时候做过这道题目，当初用的离散化的方法，不过现在USACO时限改成1秒了，那个方法可能不行了
PROB Contact	枚举，输出有点烦
PROB Stamps	一个背包问题的变形
TEXT Knapsack Problems	怎么到现在才介绍背包问题啊，前面都有好几道了
PROB Factorials	高精度可以做，但是我是去接保留了最后的6位数，一直到最后。注意只保留一位数是不行的
PROB Stringsobits	直接生成的
PROB Spinning Wheels	又是一个我没看懂题的题目，然后看了标程，原来直接枚举就行了，如此简单
PROB Feed Ratios	线性代数题目，直接把方程解出来就好了
PROB Magic Squares	比较恶心的DFS，主要是转换那个状态起来比较麻烦
PROB Sweet Butter	最短路的题目，枚举每一个点作为集合点，然后求最短路
TEXT Eulerian Tours	欧拉回路，又是一个经典的算法
PROB Riding The Fences	欧拉回路的题目
PROB Shopping Offers	DP问题，状态方程又不是我自己想的，555～
PROB Camelot	著名的亚瑟王问题，我是看了解题报告才做出来的
PROB Home on the Range	DP问题，找最大子正方形，后面还有一道是找最大子矩形的，难度大了很多
PROB A Game	动态规划，好不容易自己推出来的状态转移方程
TEXT Computational Geometry	计算几何，没看：（
PROB Closed Fences	计算几何的题目，跳过了
PROB American Heritage	二叉树遍历顺序题目，已知前序中序求后序
PROB Electric Fence	一个迭代求最优值的题目，其实就是不断缩小范围的枚举
PROB Raucous Rockers	DP，状态方程又是看来的，似乎这才是比较有难度的DP，不像前面有些题，状态方程简直显而易见

posted @ 2009-05-10 21:25 杨磊阅读(480) | 评论 (0) | 编辑收藏

USACO Section 2回顾

TEXT Graph Theory	很有用的东西，建议仔细看看
TEXT Flood Fill Algorithms	其实就是DFS
PROB The Castle	Flood Fill，直接用上面那篇文章的算法就可以过
PROB Ordered Fractions	2次循环，求出所有的分数，约分，去掉重复的，排序
PROB Sorting A Three-Valued Sequence	这题我是看的结题报告，其实就是分块来交换，首先把所有的能一次交换完成的处理掉，然后处理需要两次交换的
PROB Healthy Holsteins	忘记是贪心还是背包了……-_-!
PROB Hamming Codes	直接枚举的
TEXT Data Structures	跳过
TEXT Dynamic Programming	动态规划啦，非常有必要好好看，不过这篇文章也只是对于初学者很有用
PROB Preface Numbering	罗马数字问题，把所有可能的组合先生成出来，4，9这种，然后就是求最小表示方法
PROB Subset Sums	背包问题，这题我最开始居然没看出来……，以为是要深搜的，汗啊
PROB Runaround Numbers	直接模拟的，注意判断是否是round number的条件
PROB Party Lamps	我当初只注意到了每个操作做两次就跟没做一样，所以一共也就有8种操作，后来看了解题报告，发现其实只要处理前6个灯就可以了
PROB The Longest Prefix	DP，我看得别人的解题报告，没办法DP是我的弱项
PROB Cow Pedigrees	DP，自己推了一个差不多的状态方程，可惜错了……
PROB Zero Sum	直接模拟，把表达式生成出来，然后计算结果就行
PROB Money Systems	背包问题
PROB Controlling Companies	看了别人的解题报告，这道题目用了一个变形的Floyd算法，很巧妙
TEXT Shortest Paths	经典算法啦
PROB The Tamworth Two	模拟吧
PROB Overfencing	其实是比较恶心的一题，因为要转化那个图，剩下的就简单了，从两个exit开始BFS，然后找最大值
PROB Cow Tours	先Floyd，把图划分成两块，然后枚举
PROB Bessie Come Home	直接Floyd就行
PROB Fractions to Decimals	判断时候循环的条件就是看余数是否重复出现，当然，在我看了Analysis之后，发现了更巧妙的办法

posted @ 2009-05-10 17:43 杨磊阅读(382) | 评论 (0) | 编辑收藏

Cow XOR (USACO)

这个题目是USACO最后的一道题目，我在网上找了N多的题解，不是完全一样的，就是说的不知道是什么东西的

不知道为啥，是不是所有搞OI的人在写题解的时候都喜欢用“提示”的办法，不直接把问题说清楚，而是隐隐约约的公诉你该怎么做，然后你让你自己去想。

于是导致我到现在都没有弄明白这道题目是怎么回事，尽管他们的做法有N多种，但是归根到底都是在搞一个叫做前缀的东西。

下面这个是我在TopCoder上面找到的一个牛人的解释，算是英语写的比较好的，很通俗易懂的
上面这篇文章我想我就不用再翻译了，说的比较清楚，但是他也没有给出具体的算法，不过思路已经很清楚了

其实有了第一条性质之后，最朴素的办法就是枚举i和j，所以个2次循环，但是这样显然是要TLE的

在没有更好的算法的前提下，这道题目的算法就是空间换时间，在我看来就是这样的。

在看了N多代码之后，我觉得还是USACO的Analysis的代码看上去比较美，不知道为啥，那些用Hash和Trie的我都没看懂，可能是我比较菜吧

下面我尝试着把USACO的代码解释一下，看看能不能说清楚，很遗憾，由于这道题目的巨型的输入数据，java肯定是没办法过的

1 int main() {
2     freopen("cowxor.in", "r", stdin);
3     freopen("cowxor.out", "w", stdout);
4     scanf("%i", &n);
5     xr[0] = 0;
6     for (a = 0; a < n; a++ ) {
7         scanf("%d", &x);
8         xr[a+1] = xr[a] ^ x;
9     }
10     for (a = 0; a <= n; a++ ) {
11         prev[a][0] = a-1;
12         prev[a][1] = -1;
13         best[a] = a-1;
14     }
15     wyk = 22;
16     res = 0;
17     while (wyk--) {
18         for (a = 1; a <= n; a++ )
19             if (prev[a][0] == -1)
20                 prev[a][1] = -1;
21             else {
22                 if (xr[a] >> wyk != xr[prev[a][0]] >> wyk) {
23                     tmp[0] = prev[prev[a][0]][1];
24                     tmp[1] = prev[a][0];
25                 } else {
26                     tmp[0] = prev[a][0];
27                     tmp[1] = prev[prev[a][0]][1];
28                 }
29                 prev[a][0] = tmp[0];
30                 prev[a][1] = tmp[1];
31             }
32         fnd = false;
33         for (a = 1; a <= n; a++ )
34             if (0 <= best[a])
35                 if ((xr[a] >> wyk) % 2 != (xr[best[a]] >> wyk) % 2 ||
36                                       0 <= prev[best[a]][1])
37                     fnd = true;
38         if (fnd) {
39             res |= 1 << wyk;
40             for (a = 1; a <= n; a++ )
41                 if (0 <= best[a] &&
42                               (xr[a] >> wyk) % 2 == (xr[best[a]] >> wyk) % 2)
43                     best[a] = prev[best[a]][1];
44         }
45     }
46     for (a = 0; best[a] == -1; a++ );
47     printf("%d %d %d"n", res, best[a]+1, a);
48     return 0;
49 }

首先，6～9行，程序把从1开始到i位结束的所有的xor都求出来保存在了数组xor里面，我想这个就是为了方便后面用到xor的那个性质。

10～14行是一个初始化的过程，这个prev的定义应该可以从USACO的Analysis上面看到：

xr[pop[q][0]]'s and xr[q]'s binary representations are the same on positions e, e+1, etc., and pop[q][0] is biggest possible with 0 ≤ pop[q][0] < q. If there's no such pop[q][0] possible, then pop[q][0] = -1.

xr[pop[q][1]]'s and xr[q]'s binary representations are the same on positions e+2, e+2, etc., different on position e, and pop[q][1] is biggest possible with 0 ≤ pop[q][1] < q. If there's no such pop[q][1] possible, then pop[q][1] = -1.

我们要根据后面的循环来看这个prev数组的含义

外侧的循环的作用是每次处理一位，由于题目中已经说明了，最多只有21位，所以循环21次就可以了。

我们来看内侧的循环

18～31行，对所有的奶牛编号循环一遍，得到的结果就是：
对于当前的xor number，对于这个xor number的前21 - wyk位，与他相同的那个xor number的位置，并且，这个位置要尽量的靠后。

如果没有相同的，那么这个位置就是-1

这样，经过每次循环之后，prev里面就是与当前的xor number前k位相同的那个数的位置，一次一次循环，直到最后。

prev[i][0]存的是当current bit也相同的时候的位置，prev[i][1]存的是currnet bit不相同时候的位置，为什么要这样呢？

这可能就需要解释一下
     if (xr[a] >> wyk != xr[prev[a][0]] >> wyk) {
                     tmp[0] = prev[prev[a][0]][1];
                     tmp[1] = prev[a][0];
     } else {
                     tmp[0] = prev[a][0];
                     tmp[1] = prev[prev[a][0]][1];
    }
如果当前比较的这一位相等，那么也就意味着prev[a][0]不用变，但是prev[i][1]却必须变化，因为当前的这一位，已经不是刚才比较的那一位了，prev[a][1]应该等于此时的prev[a][0]的prev[a][1]，我想这个应该不难理解。

另一方面，如果当前比较的这一位不相等的时候，那么prev[a][1]就应该等于prev[a][0]了，因为之前的所有位都相等，之后当前这一位不相等，这就是prev[a][1]的定义。那么prev[a][0]的值呢？我们需要注意的时候，当我们处理到a的时候，prev[a][0]位置的值肯定是已经处理过的了。那么prev[a][prev[a][0]][1]里面存的就是与prev[a][0]在当前位不相等的那个xor number的位置，注意，bit只有0和1，prev[a][0]与当前的不相等，而prev[a][prev[a][0]][1]又与prev[a] [0]不相等，所以当前的prev[a][1]肯定与prev[a][prev[a][0]][1]是相等的。这就是 tmp[0] = prev[prev[a][0]][1];的含义。

我再来看32～37行，首先要知道best[q]的定义：

if X would be equal biggest possible xor, then xr[best[q]] xor xr[q]'s in equal X's binary representation on positions e, e+1, etc., and best[q] is biggest possible with 0 ≤ best[q] < q. If there's no such best[q] possible, then best[q] = -1.

想要得到尽量大的xor，那么就要尽量让每一位都不相同（xr[a] >> wyk) % 2就是取当前的二进制的最后一位

这段代码的作用就是找，到当前位为止，是否有两个xor number，他们的每一位都不相同，这样就能保证异或结果是最大的。

接下来看38～45的这段代码，因为我们是从高位到低位来处理的，这样的话，如果能保证高位是1，那么比你所有的低位都是1都管用：）

于是，既然我们找到了高位是1的情况，那么我们就要记录下结果 res

然后，剩下的那个循环就是更新best[a]

在所有的xor number中，如果当前位跟best[a]的当前位是相等的，那么就更新best[a]，将其更新为prev[best[a]][1]，就是除了当前位不相等，其余位不变的那个xor number，这样就保证了从高位开始，每一位都能够尽量取到1，因为只要高位尽量是1，我们的结果就能尽量的大。

其实prev这个数组里面真正有用的是prev[q][1]

不知道我解释清楚了没，其实解释了一遍，自己也清楚了很多。

posted @ 2009-05-10 17:20 杨磊阅读(680) | 评论 (0) | 编辑收藏

USACO Section 1回顾

Section 1.0	TEXT Introduction	介绍啦，我是没看
Section 1.1	TEXT Submitting Solutions	交你怎么提交程序的，可以看看
	PROB Your Ride Is Here	最直接的方法是直接乘，然后mod 47,不过可以利用余数定理，边乘边mod
	TEXT Contest Problem Types	跳过
	TEXT Ad Hoc Problems	跳过
	PROB Greedy Gift Givers	简单的模拟题，就是处理名字的时候有点烦
	PROB Friday the Thirteenth	数日期的题，我不知道一天天的模拟能不能过，我是只算了周五这一天的。
	PROB Broken Necklace	也是模拟题，不过很要细心，有很多特殊情况，比如全是w。
Section 1.2	TEXT Complete Search	跳过
	PROB Milking Cows	直接模拟应该是过不了的，
	PROB Transformations	模拟题，直接把所有可能的pattern生成出来，然后比较就行
	PROB Name That Number	正确方法是把字典里面的所有word转化成数字，然后比较就行。
	PROB Palindromic Squares	直接枚举
	PROB Dual Palindromes	DFS，注意搜索的时候，只要搜索回文数前一半就行，后面的直接反向复制一下就好
Section 1.3	TEXT Greedy Algorithm	跳过
	PROB Mixing Milk	简单的贪心
	PROB Barn Repair	也是贪心法，把最大的缝隙就出来，然后去覆盖
	TEXT Winning Solutions	跳过
	PROB Calf Flac	枚举，从没一点向两边枚举
	PROB Prime Cryptarithm	直接枚举，反正只有5个数
Section 1.4	TEXT More Search Techniques	跳过
	PROB Packing Rectangles	恶心题，我没做：P
	PROB The Clocks	看了一个牛人的结题报告后过的，那位牛人总结了一个数组，就是如何让表针转一圈回到原来位置的操作组合
	PROB Arithmetic Progressions	搜索，硬搜的
	PROB Mother's Milk	BFS，把所有的情况都弄出来
Section 1.5	TEXT Introduction to Binary Numbers	跳过
	PROB Number Triangles	经典DP
	PROB Prime Palindromes	搜索，生成回文数，检查是否是素数。需要一点点剪枝（长度是偶数的回文数，除了11之外必然是合数，因它肯定是11的倍数）
	PROB SuperPrime Rib	直接枚举
	PROB Checker Challenge	八皇后啊，用最经典的算法就能过，不过如果想优化的非常快，可能需要其他的办法，也有很复杂的。

posted @ 2009-05-09 15:05 杨磊阅读(1364) | 评论 (0) | 编辑收藏

终于把USACO做完了

感想及题解待会儿再发：）

posted @ 2009-05-09 13:50 杨磊阅读(336) | 评论 (0) | 编辑收藏

Picture （USACO）

1 import java.io.*;
2 import java.util.*;
3 /*
4 ID: yanglei4
5 LANG: JAVA
6 TASK:picture
7 */
8 public class picture {
9     public int[] level;
10     public int N;
11     public int ans = 0;
12
13     public class Line implements Comparable<Line> {
14         int s,t,p;
15         boolean start;
16         public Line(int a, int b, int c, boolean d) {
17             s = a;
18             t = b;
19             p = c;
20             start = d;
21         }
22         public int compareTo(Line o) {
23             if (this.p == o.p) {
24                 if (this.start)
25                     return -1;
26                 else
27                     return 1;
28             }
29             return this.p < o.p ? -1 : 1;
30         }
31     }
32     public void Scan(Line[] L) {
33         for (int i = 0;i <= 20000;i++)
34             level[i] = 0;
35         for (int i = 0; i < 2 * N;i++) {
36             if (L[i].start) {
37                 for (int j = L[i].s;j < L[i].t;j++) {
38                     level[j]++;
39                     if (level[j]==1)
40                         ans++;
41                 }
42             }
43             else {
44                 for (int j = L[i].s;j < L[i].t;j++) {
45                     level[j]--;
46                     if (level[j]==0)
47                         ans++;
48                 }
49             }
50         }
51     }
52     public void done() throws IOException {
53         BufferedReader f = new BufferedReader(new FileReader("picture.in"));
54         PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("picture.out")));
55         N = Integer.parseInt(f.readLine());
56         Line[] Lx = new Line[2 * N];
57         Line[] Ly = new Line[2 * N];
58
59         for (int i = 0; i < N; i++) {
60             StringTokenizer st = new StringTokenizer(f.readLine());
61             int x1 = Integer.parseInt(st.nextToken());//left x
62             int y1 = Integer.parseInt(st.nextToken());//left y
63             int x2 = Integer.parseInt(st.nextToken());//right x
64             int y2 = Integer.parseInt(st.nextToken());//right y
65             x1 += 10000;
66             y1 += 10000;
67             x2 += 10000;
68             y2 += 10000;
69             Lx[2 * i] = new Line(x1,x2,y1,true);
70             Lx[2 * i + 1] = new Line(x1,x2,y2,false);
71             Ly[2 * i] = new Line(y1,y2,x1,true);
72             Ly[2 * i + 1] = new Line(y1,y2,x2,false);
73         }
74         Arrays.sort(Lx);
75         Arrays.sort(Ly);
76         level = new int[20001];
77         Scan(Lx);
78         Scan(Ly);
79         out.println(ans);
80
81         out.close();
82     }
83     public static void main(String[] args) throws IOException {
84         picture t = new picture();
85         t.done();
86         System.exit(0);
87     }
88 }
89

这道题用了离散化的方法

其实最朴素的方法就是在一个20000*20000的矩阵上面把所有的点描出来，然后把这些点的周长算出来，不过算周长这一步也是很麻烦的。

因为毕竟还有可能出现“空心”的情况

用离散化的方法就是想办法只数每条线段，而不去管其他空白的地方是怎么样的。

由于我们是需要算周长的，所以只需要看矩形的四条边就可以了。

然后，我们就是先看所有的横边，再看竖边。

先把横边全部选出来，放在一个数组里面，然后排序，从下到上。

一个矩形的两条边要分成始边和终边，排序的时候，如果纵坐标相同，那么应该把始边放在终边。

如果遇到始边，那么就把这条边上面的所有点level[j]加一。

如果遇到终边，就把那条边上所有的点的level[j]减一

于是，当一条边上的点的leve是1的时候，那么就说明这条边肯定是始边边缘，所以要ans++

另一种情况，当一条边上的点的level是0的时候，那么说明这个点是终边边缘，所以ans++

这样扫完横边再扫竖边。

最后的ans就是周长了。

不过这个算法的时间效率并不是非常的好，因为数据比较弱（可能是这样）

如果真的是有5000个矩形，没个矩形都是20000×20000的，那么可能会超时

posted @ 2009-05-07 10:33 杨磊阅读(294) | 评论 (0) | 编辑收藏

USACO终于到5.5了

虽然从5.1开始，大部分题目都要借助于NoCow和网上的解题报告，但是还是学到了不少的东西。

原来认为只要熟练的掌握各种算法，那就可以随便去切题，现在发现其实不是这样。

有很多题目，都需要进行一些转化，也可以说是建模，才能套用现成的算法

而有一些题目，根本就没有现成的算法，只能你自己去想

这种算法基本上是不属于任何一类的

还有一些比如说剪枝，虽然搜索谁都会，Brute Force谁都会写，但是剪枝却不是谁都能写的出来的

这就需要一些数学功底

现在发现这个东西必须要长年累月的积累才能够达到驾轻就熟的境界。

但是就算那样，也不能保证所有的题目都会做。

posted @ 2009-05-06 23:57 杨磊阅读(287) | 评论 (0) | 编辑收藏

Telecowmunication （USACO）

  1 import java.io.*;
  2 import java.util.*;
  3 /*
  4 ID:
  5 LANG: JAVA
  6 TASK:telecow
  7 */
  8 public class telecow {
  9
10     public static final int WHITE = 0, GRAY = 1, BLACK = 2;
11     private int[][] flow, capacity, res_capacity;
12     private int[] parent, color, queue;
13     private int[] min_capacity;
14     private int size, source, sink, first, last;
15     private static int MAXN = 200;
16     int N,M;
17
18     private int maxFlow()  // Edmonds-Karp algorithm with O(V3E) complexity
19     {
20         flow = new int[MAXN][MAXN];
21         res_capacity = new int[MAXN][MAXN];
22         parent = new int[MAXN];
23         min_capacity = new int[MAXN];
24         color = new int[MAXN];
25         queue = new int[MAXN];
26         int max_flow = 0;
27         for (int i = 0; i < size; i++)
28             for (int j = 0; j < size; j++)
29                 res_capacity[i][j] = capacity[i][j];
30
31         while (BFS(source))
32         {
33             max_flow += min_capacity[sink];
34             int v = sink, u;
35             while (v != source)
36             {
37                 u = parent[v];
38                 flow[u][v] += min_capacity[sink];
39                 flow[v][u] -= min_capacity[sink];
40                 res_capacity[u][v] -= min_capacity[sink];
41                 res_capacity[v][u] += min_capacity[sink];
42                 v = u;
43             }
44         }
45         return max_flow;
46     }
47
48     private boolean BFS(int source)  // Breadth First Search in O(V2)
49     {
50         for (int i = 0; i < size; i++) {
51             color[i] = WHITE;
52             min_capacity[i] = Integer.MAX_VALUE;
53         }
54
55         first = last = 0;
56         queue[last++] = source;
57         color[source] = GRAY;
58
59         while (first != last)  // While "queue" not empty..
60         {
61             int v = queue[first++];
62             for (int u = 0; u < size; u++)
63                 if (color[u] == WHITE && res_capacity[v][u] > 0)
64                 {
65                     min_capacity[u] = Math.min(min_capacity[v], res_capacity[v][u]);
66                     parent[u] = v;
67                     color[u] = GRAY;
68                     if (u == sink) return true;
69                     queue[last++] = u;
70                 }
71         }
72         return false;
73     }
74
75     public void done() throws IOException {
76         BufferedReader f = new BufferedReader(new FileReader("telecow.in"));
77         PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("telecow.out")));
78         StringTokenizer st = new StringTokenizer(f.readLine());
79         N = Integer.parseInt(st.nextToken());
80         M = Integer.parseInt(st.nextToken());
81         source = Integer.parseInt(st.nextToken());
82         sink = Integer.parseInt(st.nextToken());
83         source--;
84         sink--;
85         int osource = source;
86         int osink = sink;
87         capacity = new int[N * 2][N * 2];
88
89         //init
90         for (int i = 0; i < N; i++)
91             capacity[i * 2][i * 2 + 1] = 1;
92
93         for (int i = 0; i < M; i++) {
94             st = new StringTokenizer(f.readLine());
95             int x = Integer.parseInt(st.nextToken());
96             int y = Integer.parseInt(st.nextToken());
97             x--;
98             y--;
99             capacity[x * 2 + 1][y * 2] = Integer.MAX_VALUE;
100             capacity[y * 2 + 1][x * 2] = Integer.MAX_VALUE;
101         }
102
103         for (int i = 0; i < 2 * N; i++) {
104             capacity[i][source * 2] = 0;
105             capacity[sink * 2 + 1][i] = 0;
106         }
107
108         int[] ans = new int[N + 1];
109
110         size = 2 * N;
111         source = source * 2 + 1;
112         sink = sink * 2;
113
114         int max = maxFlow();
115
116         int count = 0;
117         for (int i = 0; i < N; i++)
118             if (i != osource && i != osink) {
119                 capacity[i * 2][i * 2 + 1] = 0;
120                 int nowFlow = maxFlow();
121
122                 if (max - nowFlow == count + 1)
123                     ans[count++] = i;
124                 else
125                     capacity[i * 2][i * 2 + 1] = 1;
126             }
127
128         out.println(max);
129
130         for (int i = 0; i < count - 1; i++)
131             out.print(ans[i] + 1 + " ");
132         out.println(ans[count - 1] + 1);
133         out.close();
134
135     }
136
137     public static void main(String[] args) throws IOException {
138         telecow t = new telecow();
139         t.done();
140         System.exit(0);
141     }
142 }
143

这道题目我自己看出来是最小割的问题，而之前的题目有一道是最小割的

但是有一个不同，这道题是点的最小割，而不是常规的边的最小割

所以这就比较麻烦，需要我们把点转化成边

这就让我想起了之前的那个Fence Loops

我就是把所有的边表示的图转化成了点表示的图，我实在是不想再写那么恶心的算法了。

然后我就去NoCow上面看了一下，果然是有很赞的方法。

具体方法就是，把一个点变成两个点，然后在两个点之间加一条边，并且这条边的权值是1

这样的话，如果割这条边，就相当于去掉了这个点。

剩下的事情就是跟前面的那个Pollute Control差不多了

每次去掉一条边，然后用MaxFlow求一下最大流，如果得出的最大流==最大流-边的权值

那么就证明这条边是最小割。

就这样把所有的最小割找出来，然后就可以了

貌似数据很弱，不像上次的那个Pollute那道题目，还要考虑边的编号的问题的。

posted @ 2009-05-06 14:01 杨磊阅读(267) | 评论 (0) | 编辑收藏

All Latin Squares (USACO)

一个搜索的题目，不过肯定是要剪枝的

最简单的两个剪枝我想到了

第一个：
首先，第一行已经确定了，那么我们可以把第一列也确定，就按照升序，2，3，4，5……，这样的话，没生成这么一个square，我们就可以随便的去交换除了第一行后面的几行。

他们的一个全排列就对应着一种组合，所以最后的答案乘以N-1的阶乘就可以

第二个：
这个其实是看来的，就是每次只要所搜完N-1行就可以了。因为剩下的一行必然存在一个解。其实这个不难理解的，最后一行的排列顺序只跟前面的每一列缺少的那个数有关。

而且也只能取缺少的那个数，所以也就是唯一的。

第三个：
要用到置换群，我没看懂

尽管之前N次看组合数学里面都说到置换群，但是我还是似懂非懂，问了数学小王子，他也不是非常懂。然后以为这个东西没用，就忽略了。没想到还真的有用。

posted @ 2009-05-02 15:22 杨磊阅读(390) | 评论 (0) | 编辑收藏

Network of Schools （USACO)

今天终于第一次写了强连通分量的题目

虽然以前老早就知道有这么个东西，对这个东西也有概念，但是确实从来没有写过判别的算法

原来如此简单，但是确实也很巧妙。

在算法导论上面看到的K算法，方法就是做两遍DFS，因为强连通分量有一个性质，就是如果把所有的边反向，那么它仍然是一个强连通分量。

但是今天我用的是Tarjan算法，据说效率比K算法高很多，事实上也应该是这样，因为Tarjan算法只做了一次DFS

其实思想也很简单，就是一直DFS（u），向下向下向下，如果突然发现有一个点可以回到u了，那么这个肯定是一个强连通分量。

哈哈，是不是很通俗。

严格的定义过程大家可以看这里http://www.cmykrgb123.cn/blog/scc-tarjan/

我也是参考了这个才做出来的Tarjan算法，Wiki上的那个过于庞大了，虽然封装的非常好

说说本题，其实就是找强连通分量，然后把每个强连通分量当成一个“超点”来考虑

如果这个“超点”的入度为零，那么它就必然是一个源头，统计这样的“超点”的个数，就是第一问的答案。

第二问，如果有一个“超点”入度不是0，但是出度是0，那就说明这个“超点”可以给其他“超点”作贡献。

我们就找这样的点对，把入度是0和出度是0的点连起来。

这样匹配过后，剩下的要么全是入度是0的，要么全是出度是0的，这些就没办法了，随便从一个连通分量连接过来一条边就可以了。

所以第二问的答案就是所有的“超点”中，入度是0和出度是0的点大的那个数。

  1 import java.io.*;
  2 import java.util.*;
  3 /*
  4 ID: yanglei4
  5 LANG: JAVA
  6 TASK:schlnet
  7 */
  8 public class schlnet {
  9     boolean[] inStack;
10     int[] DFN;
11     int[] LOW;
12     int index = 0;
13     int[] Stack;
14     int top = 0;
15     int[][] map;
16     int BelongCount = 0;
17     int[] belong;
18
19     public void Tarjan(int i) {
20         DFN[i] = LOW[i] = ++index;
21         inStack[i] = true;
22         Stack[top++] = i;
23         for (int j = 1; j <= map[i][0]; j++) {
24             if (DFN[map[i][j]] == 0) {
25                 Tarjan(map[i][j]);
26                 if (LOW[map[i][j]] < LOW[i])
27                     LOW[i] = LOW[map[i][j]];
28             }
29             else if (inStack[map[i][j]] && DFN[map[i][j]] < LOW[i])
30                 LOW[i] = DFN[map[i][j]];
31         }
32         if (DFN[i] == LOW[i]) {
33             int j = 0;
34             do {
35                 j = Stack[--top];
36                 inStack[j] = false;
37                 belong[j] = BelongCount;
38             } while (j != i);
39             BelongCount++;
40         }
41
42     }
43
44     public void done() throws IOException {
45         BufferedReader f = new BufferedReader(new FileReader("schlnet.in"));
46         PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("schlnet.out")));
47         int N = Integer.parseInt(f.readLine());
48         map = new int[N + 1][N + 1];
49         DFN = new int[N + 1];
50         LOW = new int[N + 1];
51         inStack = new boolean[N + 1];
52         Stack = new int[N + 1];
53         belong = new int[N + 1];
54
55         Arrays.fill(belong,-1);
56         for (int i = 1; i <= N; i++) {
57             StringTokenizer st = new StringTokenizer(f.readLine());
58             int len = st.countTokens();
59             map[i][0] = len - 1;
60             for (int j = 1; j <= len - 1; j++)
61                 map[i][j] = Integer.parseInt(st.nextToken());
62         }
63
64         for (int i = 1; i <= N; i++) {
65             if (DFN[i] == 0)
66                 Tarjan(i);
67         }
68         boolean[][] dis = new boolean[BelongCount + 1][BelongCount + 1];
69         for (int i = 1;i <= N;i++) {
70             for (int k = 1;k <= map[i][0];k++)
71                 dis[belong[i] + 1][belong[map[i][k]] + 1] = true;
72         }
73         int[] oud = new int[BelongCount + 1];
74         int[] ind = new int[BelongCount + 1];
75
76         for (int i = 1;i <= BelongCount;i++) {
77             for (int j = 1; j<= BelongCount;j++) {
78                 if (i!=j && dis[i][j]) {
79                     oud[i]++;
80                     ind[j]++;
81                 }
82             }
83         }
84
85         int i0 = 0;
86         int o0 = 0;
87         for (int i = 1;i <= BelongCount;i++) {
88             if (ind[i] == 0)
89                 i0++;
90             if (oud[i] == 0)
91                 o0++;
92         }
93
94         if (BelongCount == 1) {
95             out.println("1");
96             out.println("0");
97         }
98         else {
99             out.println(i0);
100             out.println(i0>o0?i0:o0);
101         }
102         out.close();
103     }
104
105     public static void main(String[] args) throws IOException {
106         schlnet t = new schlnet();
107         t.done();
108         System.exit(0);
109     }
110 }
111

posted @ 2009-04-30 20:23 杨磊阅读(274) | 评论 (0) | 编辑收藏

Big Barn （USACO）

1 import java.io.*;
2 import java.util.*;
3
4 /*
5 ID: yanglei4
6 LANG: JAVA
7 TASK:bigbrn
8 */
9 public class bigbrn {
10     public static int min(int a, int b) {
11         if (a < b) return a;
12         else return b;
13     }
14     public static int min3(int a, int b, int c) {
15         return min(a, min(b, c));
16     }
17     public static void main(String[] args) throws IOException {
18         BufferedReader f = new BufferedReader(new FileReader("bigbrn.in"));
19         PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("bigbrn.out")));
20         StringTokenizer st = new StringTokenizer(f.readLine());
21         int N = Integer.parseInt(st.nextToken());
22         int T = Integer.parseInt(st.nextToken());
23         int[][] map = new int[N][N];
24         for (int i = 0; i < T; i++) {
25             st = new StringTokenizer(f.readLine());
26             int x = Integer.parseInt(st.nextToken());
27             int y = Integer.parseInt(st.nextToken());
28             map[x - 1][y - 1] = -1;
29         }
30
31         for (int i = 0; i < N; i++) {
32             if (map[0][i]!= -1)
33                 map[0][i] = 1;
34             if (map[i][0]!= -1)
35                 map[i][0] = 1;
36         }
37
38
39         for (int i = 1; i < N; i++)
40             for (int j = 1; j < N; j++) {
41                 if (map[i][j] != -1) {
42                     int temp = min3(map[i - 1][j], map[i][j - 1], map[i - 1][j - 1]);
43                     if (temp != -1) map[i][j] = temp + 1;
44                     else map[i][j] = 1;
45                 }
46             }
47
48         int max = 0;
49         for (int i = 0; i < N; i++)
50             for (int j = 0; j < N; j++)
51                 if (map[i][j] != 0 && map[i][j] > max)
52                     max = map[i][j];
53
54         out.println(max);
55         out.close();
56         System.exit(0);
57     }
58 }
59

应该算是一个比较基本的DP吧，状态转移方程也不难想，但是我最开始写成了N^3的了

首先就是用Map[i][j]来表示以i，j为右下角的最大的square的大小

初始化就是第一行，第一列，如果不是#，那么肯定是1

然后对于i，j，我们需要观察i - 1，j - 1，因为是square，所以只跟这个有关

如果在第i行，第j列上面，从当前位置开始，连续map[i-1][j-1]个位置，没有#的话，那么map[i][j] = map[i-1][j-1]+1

我就是在判断连续map[i-1][j-1]这个地方出了问题，我又加了一层循环，所以就变成了N^3的了，然后果然TLE了

这个地方完全没必要用循环一个一个去判断，因为其实你已经有结果了，这个结果就是map[i-1][j]和map[i][j-1]里面小的那个

map[i-1][j]肯定就是从当前位置开始，在第j列上，向上最多可以走多少步不碰到#

因为这时候实际上你已经确定了，#只有可能出现在第i行，第j列上，因为map[i-1][j-1]不是#就保证了这一点

于是，找到两个方向上走的比较近的那个数，如果这个数是小于map[i-1][j-1]的，那么map[i][j]就等于这个数

否则，map[i][j] = map[i-1][j-1]+1

这个地方的重点就是，如果map[i-1][j-1]不是#，那么就保证了#只能在第i行，第j列上面。

只需要检查这两个就可以

然后我们就可以来看map[i-1][j],map[i][j-1]，这两个东西其实跟map[i-1][j-1]共享了上面的一块。

如果在第i行，第j列上面出现了#，那么map[i-1][j],map[i][j-1]肯定比map[i-1][j-1]小

否则，我们的square最大也就只能是map[i-1][j-1]+1，因为map[i-1][j-1]已经是以i-1,j-1为右下角最大的square了

于是状态转移方程就是

map[i][j] = min (map[i-1][j],map[i][j-1],map[i-1][j-1]) + 1, map[i][j] != '#'

posted @ 2009-04-30 10:40 杨磊阅读(176) | 评论 (0) | 编辑收藏

Starry Night (USACO)

摘要: 绝对，非常考耐心，细心的一道题目，这道题目充分的说明我是考虑问题不全面的人所以现在看来TopCoder的测试数据还都算是比较弱的，真的没有极其变态的，而且TopCoder的题目其实没有超级复杂的，或许是因为我从来没做过第三题吧。回正题。这道题目其实一看就知道怎么做，但是就一个字烦首先你要用FloodFill把所有的Cluster标注出来，这个不难，而且速度很快，时间... 阅读全文

posted @ 2009-04-28 21:53 杨磊阅读(476) | 评论 (2) | 编辑收藏

终于到了5.1了，希望能早日做完USACO

发个网络流的标准代码，用的是BFS

外面自己套一个类就行了

capacity自己初始化放进去就可以了

返回值在max_flow里面，有需要的话可以自己改成返回类型的

1     public static final int WHITE = 0, GRAY = 1, BLACK = 2;
2     private int[][] flow, capacity, res_capacity;
3     private int[] parent, color, queue;
4     private int[] min_capacity;
5     private int size, source, sink, first, last;
6     private int max_flow;
7
8     private void maxFlow()  // Edmonds-Karp algorithm with O(V3E) complexity
9     {
10         flow = new int[size][size];
11         res_capacity = new int[size][size];
12         parent = new int[size];
13         min_capacity = new int[size];
14         color = new int[size];
15         queue = new int[size];
16
17         for (int i = 0; i < size; i++)
18             for (int j = 0; j < size; j++)
19                 res_capacity[i][j] = capacity[i][j];
20
21         while (BFS(source))
22         {
23             max_flow += min_capacity[sink];
24             int v = sink, u;
25             while (v != source)
26             {
27                 u = parent[v];
28                 flow[u][v] += min_capacity[sink];
29                 flow[v][u] -= min_capacity[sink];
30                 res_capacity[u][v] -= min_capacity[sink];
31                 res_capacity[v][u] += min_capacity[sink];
32                 v = u;
33             }
34         }
35     }
36
37     private boolean BFS(int source)  // Breadth First Search in O(V2)
38     {
39         for (int i = 0; i < size; i++)
40         {
41             color[i] = WHITE;
42             min_capacity[i] = Integer.MAX_VALUE;
43         }
44
45         first = last = 0;
46         queue[last++] = source;
47         color[source] = GRAY;
48
49         while (first != last)  // While "queue" not empty..
50         {
51             int v = queue[first++];
52             for (int u = 0; u < size; u++)
53                 if (color[u] == WHITE && res_capacity[v][u] > 0)
54                 {
55                     min_capacity[u] = Math.min(min_capacity[v], res_capacity[v][u]);
56                     parent[u] = v;
57                     color[u] = GRAY;
58                     if (u == sink) return true;
59                     queue[last++] = u;
60                 }
61         }
62         return false;
63     }

posted @ 2009-04-28 10:39 杨磊阅读(182) | 评论 (0) | 编辑收藏

Frame Up

原来这道题目这么简单，记得当年高中的时候做USACO的时候，好像最后就是卡在这道题目上面，然后就是NOIP了

然后我就伤心的告别了NOIP投向了好好学习的怀抱。

今天拿过来看还是没什么思路，然后就去搜了一下解题报告。

其实这个题目无论如何也是要DFS的，因为人家要你输出全部的答案。这样就不用怕了，最多就是剪剪枝。

但是最开始没有想到这一点。

问题就在这个剪枝上，和怎么判断合理的解上面。

方法就是，首先找到每一个Frame的四个角。

然后沿着这个Frame的每条边走一下，如果有其他的字符，就是跟当前Frame不同的字符，那么这个字符肯定是在当前这个字符上层的。

这样就能用一张表来确定上下关系，Above[i][j]，表示第i个字符在第j个字符上层

然后就是根据这张表来做一个DFS，这样就可以了。

看了leokan的解题报告，说还要floyd一下。

这样做的目的无非也就是想确定两两之间的上下层关系罢了。

但是似乎没这个必要，直接DFS就可以了。

  1 import java.io.*;
  2 import java.util.*;
  3 /*
  4 ID: yanglei4
  5 LANG: JAVA
  6 TASK:frameup
  7 */
  8 public class frameup{
  9     static char[][] board;
10     static int[][] pos;
11     static int[] in;
12     static int cnt;
13     static int[][] above;
14     static char[] answer;
15     static PrintWriter out;
16     public static void findAnswer(int step) {
17         int i,j;
18         if (step >= cnt) {
19             for (int k = 0; k < cnt; k++)
20                 out.print(answer[k]);
21             out.println();
22         }
23         for (i = 0; i < 26; i++)
24         if (in[i]> 0) {
25             for (j = 0; j < 26; j++)
26             if (in[j] > 0 && above[i][j] == 1) break;
27
28               if (j >= 26) { /* no such frame, so this COULD be the next one */
29                 answer[step] = (char) ('A' + i);
30                 in[i] = 0; /* mark as in answer */
31                 findAnswer(step + 1);
32                 in[i] = 1; /* clear mark */
33               }
34
35         }
36     }
37     public static void main(String[] args) throws IOException {
38         BufferedReader f = new BufferedReader(new FileReader("frameup.in"));
39         out = new PrintWriter(new BufferedWriter(new FileWriter("frameup.out")));
40         StringTokenizer st = new StringTokenizer(f.readLine());
41         int H = Integer.parseInt(st.nextToken());
42         int W = Integer.parseInt(st.nextToken());
43
44         board = new char[30][32];
45         pos = new int[26][4];
46         in = new int[26];
47         cnt = 0;
48
49         above = new int[26][26];
50         answer = new char[27];
51
52         for (int i = 0; i < H; i++) {
53             String temp = f.readLine();
54             board[i] = temp.toCharArray();
55             for (int j = 0; j < W; j++) {
56                 if (board[i][j] != '.') {
57                     int loc = board[i][j] - 'A';
58
59                     if (in[loc] == 0) {//First time met
60                         in[loc] = 1;
61                         cnt++;
62                         pos[loc][0] = pos[loc][2] = i;
63                         pos[loc][1] = pos[loc][3] = j;
64                     }
65                     else {
66                         if (i < pos[loc][0]) pos[loc][0] = i;
67                         if (i > pos[loc][2]) pos[loc][2] = i;
68                         if (j < pos[loc][1]) pos[loc][1] = j;
69                         if (j > pos[loc][3]) pos[loc][3] = j;
70                     }
71
72                 }
73             }
74         }
75
76         for (int i = 0; i < 26; i++)
77             if (in[i] > 0) { /* for each frame */
78                  for (int j = pos[i][0]; j <= pos[i][2]; j++) { /* check left and right borders */
79
80                    /* left */
81                    int loc = board[j][pos[i][1]] - 'A';
82                    if (loc != i) /* there's a different frame where this one should be */
83                      above[loc][i] = 1; /* that different frame must be above this one */
84
85                    /* right */
86                    loc = board[j][pos[i][3]] - 'A';
87                    if (loc != i) /* same as left */
88                      above[loc][i] = 1;
89                  }
90                  for (int j = pos[i][1]; j <= pos[i][3]; j++) { /* check top and bottom */
91
92                    /* top */
93                    int loc = board[pos[i][0]][j] - 'A';
94                    if (loc != i)
95                      above[loc][i] = 1;
96
97                    /* bottom */
98                    loc = board[pos[i][2]][j] - 'A';
99                    if (loc != i)
100                      above[loc][i] = 1;
101                  }
102             }
103
104         findAnswer(0);
105         out.close();
106         System.exit(0);
107     }
108 }
109

posted @ 2009-04-27 17:07 杨磊阅读(154) | 评论 (0) | 编辑收藏

UML读书笔记

Principles of Modeling

First, The choice of what models to create has a profound influence on how a problem is attacked and how a solution is shaped.
Second, Every model may be expressed at different levels of precision.
Third, The best models are connected to reality.
Fourth,No single model or view is sufficient. Every nontrivial system is best approached through a small set of nearly independent models with multiple viewpoints.

Three major elements:

the UML's basic building blocks
the rules that dictate how those building blocks may be put together
some common mechanisms that apply throughout the UML

The vocabulary of the UML encompasses three kinds of building blocks:

Things
Relationships
Diagrams

Three kinds of relationships that are most important: dependencies, generalizations, and associations.

Dependencies are using relationships. For example, pipes depend on the water heater to heat the water they carry.
Associations are structural relationships among instances. For example, rooms consist of walls and other things; walls themselves may have embedded doors and windows; pipes may pass through walls.
Generalizations connect generalized classes to more-specialized ones in what is known as subclass/superclass or child/parent relationships. For example, a picture window is a kind of window with very large, fixed panes; a patio window is a kind of window with panes that open side to side.

posted @ 2009-04-26 18:37 杨磊阅读(170) | 评论 (0) | 编辑收藏

解决Windows 7支付宝问题

先自动下载了新的控件

然后去用这个文件去注册掉那个什么什么CAB

http://www.blogjava.net/Files/LittleDS/800A138F.rar

然后再去下载 xenroll.dll
去 cmd 运行 regsvr32 xenroll.dll

然后就可以了，就可以导入数字证书了。

posted @ 2009-04-25 14:16 杨磊阅读(163) | 评论 (0) | 编辑收藏

The Primes （USACO）

摘要: Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ --> 1 import java.io.*; 2 import java.util.*; &n... 阅读全文

posted @ 2009-04-24 12:01 杨磊阅读(208) | 评论 (0) | 编辑收藏

就一个toString和toByteArray的差别

速度提高了一倍啊，看来底层的东西还是有必要了解的，不过Java写东西是真方便啊

本来这道题是要高精度的，YD啊，但是Java有BigDecimal，哈哈，不过这玩意好用是好用，但是确实是非常的慢

我估计我自己用一个数组实现的话，速度应该还会快的

不过最后被我瞎猫碰死耗子碰到了，终于过了。

1 import java.io.*;
2 import java.math.BigInteger;
3 import java.util.*;
4 /*
5 ID: yanglei4
6 LANG: JAVA
7 TASK:buylow
8 */
9 public class buylow{
10     public static void main(String[] args) throws IOException {
11         BufferedReader f = new BufferedReader(new FileReader("buylow.in"));
12         PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("buylow.out")));
13         int N = Integer.parseInt(f.readLine());
14         int[] D = new int[N];
15         int count = 0;
16         String temp = f.readLine();
17         while (temp != null) {
18             StringTokenizer st = new StringTokenizer(temp);
19             int len = st.countTokens();
20             for (int i = 0; i < len; i++)
21                 D[count++] = Integer.parseInt(st.nextToken());
22             temp = f.readLine();
23         }
24         long[] lds = new long[N];
25         BigInteger[] cnt = new BigInteger[N];
26         for (int i = 0; i < N; i++) {
27                 lds[i] = 1;
28                 cnt[i] = new BigInteger("1");
29         }
30         ArrayList<Integer> exist = new ArrayList<Integer>();
31         for (int i = 0; i < N; i++) {
32
33             for (int j = i - 1; j >= 0; j--) {
34                 if (D[i] < D[j]) {
35                     if (lds[j] + 1 > lds[i]) {
36                         lds[i] = lds[j] + 1;
37                         exist.clear();
38                         exist.add(D[j]);
39                         cnt[i] = new BigInteger(cnt[j].toByteArray());
40                     }
41                     else if (lds[j] + 1 == lds[i]) {
42                         if (!exist.contains(D[j])) {
43                             exist.add(D[j]);
44                             cnt[i] = cnt[i].add(cnt[j]);
45                         }
46                     }
47
48                 }
49             }
50         }
51
52 /*        for (int i = 0; i < N; i++)
53             System.out.println(lds[i][0] + " " + lds[i][1] + " " + cnt[i]);*/
54
55         long maxleng = 0;
56         for (int i = 0; i < N; i++) {
57             if (maxleng < lds[i])
58                     maxleng = lds[i];
59         }
60         BigInteger sum = new BigInteger("0");
61         exist = new ArrayList<Integer>();
62         for (int i = N - 1; i >= 0; i--) {
63             if (lds[i] == maxleng)
64                 if (!exist.contains(D[i])) {
65                     sum = sum.add(cnt[i]);
66                     exist.add(D[i]);
67                 }
68         }
69
70         out.println(maxleng + " " + sum);
71         out.close();
72         System.exit(0);
73     }
74 }
75

就是红色的那个地方，用toString的话，最后一个点过不了，换成了这个就过了。

posted @ 2009-04-23 23:27 杨磊阅读(2367) | 评论 (0) | 编辑收藏

Network Flow

1   if (source = sink)
2     totalflow = Infinity
3     DONE

4   totalflow = 0

5   while (True)
6 # find path with highest capacity from
   # source to sink
7 # uses a modified djikstra's algorithm
8     for all nodes i
9       prevnode(i) = nil
10       flow(i) = 0
11       visited(i) = False
12     flow(source) = infinity

13     while (True)
14       maxflow = 0
15       maxloc = nil
16       # find the unvisited node with
         # the highest capacity to it
17       for all nodes i
18         if (flow(i) > maxflow AND
                          not visited(i))
19           maxflow = flow(i)
20           maxloc = i
21       if (maxloc = nil)
22         break inner while loop
23       if (maxloc = sink)
24         break inner while loop
24a      visited(maxloc) = true
25       # update its neighbors
26       for all neighbors i of maxloc
27         if (flow(i) < min(maxflow,
                      capacity(maxloc,i)))
28           prevnode(i) = maxloc
29           flow(i) = min(maxflow,
                    capacity(maxloc,i))

30     if (maxloc = nil)         # no path
31       break outer while loop

32     pathcapacity = flow(sink)
33     totalflow = totalflow + pathcapacity

   # add that flow to the network,
   # update capacity appropriately
35     curnode = sink
         # for each arc, prevnode(curnode),
         # curnode on path:
36     while (curnode != source)
38       nextnode = prevnode(curnode)
39       capacity(nextnode,curnode) =
           capacity(nextnode,curnode) -
40                           pathcapacity
41       capacity(curnode,nextnode) =
           capacity(curnode,nextnode) +
42                           pathcapacity
43       curnode = nextnode

posted @ 2009-04-22 14:05 杨磊阅读(155) | 评论 (0) | 编辑收藏

我的Eclipse配色

http://www.blogjava.net/Files/LittleDS/eclipse.rar 喜欢的可以拿去用，欢迎提意见

之所以用黑色，就是因为foreground可以用更多的颜色，对比度高一点

而且经常看白色的背景色，好像对眼睛也是一种折磨。

http://www.blogjava.net/Files/LittleDS/eclipse.rar

posted @ 2009-04-17 22:38 杨磊阅读(1785) | 评论 (0) | 编辑收藏

Hash很强大

第一次用Java代码过搜索+剪枝的题目啊，不容易啊……，而且还是参考了后面的Analysis，-_-!

不过里面的那个Hash的剪枝还是很强大的，再一次说明了学好数学的重要

同时，再一次证明了Java有多慢，C++肯定不用加最后的剪枝就过了，但是Java就不行

而且有一个点居然要1.5秒，真是不可思议。

待会儿我再试试Analysis里面的代码，看看有多快，是不是Static的要比正常的写快。

算法没啥好说的，就是生成，把F+R个数生成出来，然后用那个最大ratio是最小ratio的三倍做一个剪枝

我之前还打了一个表，所有的数的ratio的表，然后后面直接访问这个表，而不是去现去除

但是发现好像左右不大。

剩下的就是那个Hash的优化了，很强大，就好象题目里说的那样，就是看F生成好之后的比率

比如1 3 5 7，前两个是F的，后两个是R的，那么它的variance和2 6 10 14是一样的，这个不用我解释吧

这样的话呢，同一个比率的就不用再做第二次了。

自己没有去严格的证明这个剪枝的正确性，但是想想证明应该不太难。

同一个ratio的话肯定取前面那组，如果是不同的ratio呢？

现在的问题就是，如果是不同的ratio，后面的正确的组会不会被前面的不小心剪掉。或者后面的这组根本没必要。

如果是根本没必要的话，那么这个剪枝肯定就是正确的了。

其实这个剪枝是正确的，因为在你用比如1，3的时候，你就试过了后面所有的不同组合了。

那么当你扩大1，3到2，6时，我们可以看看USACO后面的解释的那个例子来。

39/16 = 2.43750000000000000000

40/16 = 2.50000000000000000000

39/15 = 2.60000000000000000000

40/15 = 2.66666666666666666666

39/14 = 2.78571428571428571428

40/14 = 2.85714285714285714285

39/13 = 3.00000000000000000000

40/13 = 3.07692307692307692307

39/12 = 3.25000000000000000000

40/12 = 3.33333333333333333333

就相当于这些ratio全都翻了一倍，那样的话，variance当然不会变

所以个剪枝是正确的。

贴一下代码：

  1 import java.io.*;
  2 import java.util.*;
  3 /*
  4 ID: yanglei4
  5 LANG: JAVA
  6 TASK:cowcycle
  7 */
  8 public class cowcycle{
  9     public double[][] ratio = new double[81][41];
10     int F,R,F1,F2,R1,R2;
11     public int[] s1;
12     public int[] s2;
13     double min = 1000000;
14     static String[] hash;
15     public int[] ans1;
16     public int[] ans2;
17
18     public static boolean contains(String str) {
19         int num = str.hashCode();
20         if(num<0) num = -num;
21         int p = num % hash.length;
22         while(hash[p]!=null)
23             if(str.equals(hash[p]))   return true;
24             else if(p==hash.length-1) p=0;
25             else p++;
26         hash[p]=str;
27         return false;
28    }
29
30     public void DFS_F(int step,int start) {
31         if (step == F + 1) {
32             StringBuffer str = new StringBuffer();
33             for(int i = 2;i <= F;i++)
34                 str.append((int)(100*(double)s1[i]/s1[1]));
35             if(contains(str.toString())) return;
36
37             DFS_R(1,R1);
38             return;
39         }
40         for (int i = start; i <= F2 - (F - step); i++) {
41             s1[step] = i;
42             DFS_F(step + 1, i + 1);
43         }
44     }
45
46     public void DFS_R(int step,int start) {
47         if (step == R + 1) {
48             if (s1[F] * s2[R] < 3 * s1[1] * s2[1])
49                 return;
50             count();
51             return;
52         }
53         for (int i = start; i <= R2 - (R - step); i++) {
54             s2[step] = i;
55             DFS_R(step + 1, i + 1);
56         }
57     }
58
59     public double count() {
60         double[] rate = new double[R * F + 1];
61         double[] diff = new double[R * F + 1];
62         double sum = 0;
63         double sumf = 0;
64         int l = 1;
65         for (int i = 1; i <= F; i++)
66             for (int j = 1; j <= R; j++)
67                 rate[l++] = ratio[s1[i]][s2[j]];
68
69         Arrays.sort(rate);
70
71         for (int i = 1; i <= F * R - 1; i++) {
72             diff[i] = rate[i + 1] - rate[i];
73             sum += diff[i];
74             sumf += diff[i] * diff[i];
75         }
76         double avg = sum / (F * R - 1);
77         double vf = sumf - sum * avg;
78         if (vf < min) {
79             min = vf;
80             for (int i = 1; i <= F; i++) ans1[i] = s1[i];
81             for (int i = 1; i <= R; i++) ans2[i] = s2[i];
82         }
83         return 0.0;
84     }
85
86     public void done() throws IOException {
87         for (int i = 25; i <= 80; i++)
88             for (int j = 5; j <= 40; j++)
89                 ratio[i][j] = (double)i / j;
90
91         BufferedReader f = new BufferedReader(new FileReader("cowcycle.in"));
92         PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("cowcycle.out")));
93         StringTokenizer st = new StringTokenizer(f.readLine());
94         F = Integer.parseInt(st.nextToken());
95         R = Integer.parseInt(st.nextToken());
96         st = new StringTokenizer(f.readLine());
97         F1 = Integer.parseInt(st.nextToken());
98         F2 = Integer.parseInt(st.nextToken());
99         R1 = Integer.parseInt(st.nextToken());
100         R2 = Integer.parseInt(st.nextToken());
101         s1 = new int[F + 1];
102         s2 = new int[R + 1];
103         ans1 = new int[F + 1];
104         ans2 = new int[R + 1];
105         hash = new String[50003];
106
107         DFS_F(1,F1);
108
109         for (int i = 1; i <= F - 1; i++)
110             out.print(ans1[i] + " ");
111         out.println(ans1[F]);
112
113         for (int i = 1; i <= R - 1; i++)
114             out.print(ans2[i] + " ");
115         out.println(ans2[R]);
116
117         out.close();
118     }
119
120     public static void main(String[] args) throws IOException {
121         cowcycle t = new cowcycle();
122         t.done();
123         System.exit(0);
124     }
125 }
126

posted @ 2009-04-17 22:32 杨磊阅读(130) | 评论 (0) | 编辑收藏

Cryptcowgraphy

不明白为什么一模一样的算法，Java的结果就是不对呢，很奇怪

又是一个剪枝的题目，我发现剪枝这东西真是需要灵感的

那个不会改变的字符串的剪枝比较容易想，不过Hash的那个剪枝确实是让我爱莫能及啊，然后就参考了Leokan的代码

很神奇的代码，Hash Size只有3W+，然后居然就AC了，我最后就是用他的代码试了一下 -_-!，然后AC的

可是为什么我的Java代码就是过不了呢？很神奇

在第七个点的时候，Hash Size要开到50W才可以出正确的结果，但是这样的话时间会超长，估计有4，5秒的样子

而且我把能加的剪枝全加上了，还是不行……

真是不明白所以啊，不知道自己问题处在哪里了。

太TM诡异了……

posted @ 2009-04-16 23:56 杨磊阅读(150) | 评论 (0) | 编辑收藏

转一篇字符串Hash函数的文章

http://www.cmykrgb123.cn/blog/string-hash-compare/

常用的字符串Hash函数还有ELFHash，APHash等等，都是十分简单有效的方法。这些函数使用位运算使得每一个字符都对最后的函数值产生影响。另外还有以MD5和SHA1为代表的杂凑函数，这些函数几乎不可能找到碰撞。

常用字符串哈希函数有BKDRHash，APHash，DJBHash，JSHash，RSHash，SDBMHash，PJWHash，ELFHash等等。对于以上几种哈希函数，我对其进行了一个小小的评测。

Hash函数	数据1	数据2	数据3	数据4	数据1得分	数据2得分	数据3得分	数据4得分	平均分
BKDRHash	2	0	4774	481	96.55	100	90.95	82.05	92.64
APHash	2	3	4754	493	96.55	88.46	100	51.28	86.28
DJBHash	2	2	4975	474	96.55	92.31	0	100	83.43
JSHash	1	4	4761	506	100	84.62	96.83	17.95	81.94
RSHash	1	0	4861	505	100	100	51.58	20.51	75.96
SDBMHash	3	2	4849	504	93.1	92.31	57.01	23.08	72.41
PJWHash	30	26	4878	513	0	0	43.89	0	21.95
ELFHash	30	26	4878	513	0	0	43.89	0	21.95

其中数据1为100000个字母和数字组成的随机串哈希冲突个数。数据2为100000个有意义的英文句子哈希冲突个数。数据3为数据1的哈希值与 1000003(大素数)求模后存储到线性表中冲突的个数。数据4为数据1的哈希值与10000019(更大素数)求模后存储到线性表中冲突的个数。

经过比较，得出以上平均得分。平均数为平方平均数。可以发现，BKDRHash无论是在实际效果还是编码实现中，效果都是最突出的。APHash也是较为优秀的算法。DJBHash,JSHash,RSHash与SDBMHash各有千秋。PJWHash与ELFHash效果最差，但得分相似，其算法本质是相似的。

在信息修竞赛中，要本着易于编码调试的原则，个人认为BKDRHash是最适合记忆和使用的。。

附：各种哈希函数的C语言程序代码

unsigned int SDBMHash(char *str)
{
unsigned int hash = 0;
while (*str)
{
// equivalent to: hash = 65599*hash + (*str++);
hash = (*str++) + (hash << 6) + (hash << 16) - hash;
}
return (hash & 0x7FFFFFFF);
}

// RS Hash Function
unsigned int RSHash(char *str)
{
unsigned int b = 378551;
unsigned int a = 63689;
unsigned int hash = 0;
while (*str)
{
hash = hash * a + (*str++);
a *= b;
}
return (hash & 0x7FFFFFFF);
}

// JS Hash Function
unsigned int JSHash(char *str)
{
unsigned int hash = 1315423911;
while (*str)
{
hash ^= ((hash << 5) + (*str++) + (hash >> 2));
}
return (hash & 0x7FFFFFFF);
}

// P. J. Weinberger Hash Function
unsigned int PJWHash(char *str)
{
unsigned int BitsInUnignedInt = (unsigned int)(sizeof(unsigned int) * 8);
unsigned int ThreeQuarters    = (unsigned int)((BitsInUnignedInt  * 3) / 4);
unsigned int OneEighth        = (unsigned int)(BitsInUnignedInt / 8);
unsigned int HighBits         = (unsigned int)(0xFFFFFFFF) << (BitsInUnignedInt - OneEighth);
unsigned int hash             = 0;
unsigned int test             = 0;
while (*str)
{
hash = (hash << OneEighth) + (*str++);
if ((test = hash & HighBits) != 0)
{
hash = ((hash ^ (test >> ThreeQuarters)) & (~HighBits));
}
}
return (hash & 0x7FFFFFFF);
}

// ELF Hash Function
unsigned int ELFHash(char *str)
{
unsigned int hash = 0;
unsigned int x    = 0;
while (*str)
{
hash = (hash << 4) + (*str++);
if ((x = hash & 0xF0000000L) != 0)
{
hash ^= (x >> 24);
hash &= ~x;
}
}
return (hash & 0x7FFFFFFF);
}

// BKDR Hash Function
unsigned int BKDRHash(char *str)
{
unsigned int seed = 131; // 31 131 1313 13131 131313 etc..
unsigned int hash = 0;
while (*str)
{
hash = hash * seed + (*str++);
}
return (hash & 0x7FFFFFFF);
}

// DJB Hash Function
unsigned int DJBHash(char *str)
{
unsigned int hash = 5381;
while (*str)
{
hash += (hash << 5) + (*str++);
}
return (hash & 0x7FFFFFFF);
}

// AP Hash Function
unsigned int APHash(char *str)
{
unsigned int hash = 0;
int i;
for (i=0; *str; i++)
{
if ((i & 1) == 0)
{
hash ^= ((hash << 7) ^ (*str++) ^ (hash >> 3));
}
else
{
hash ^= (~((hash << 11) ^ (*str++) ^ (hash >> 5)));
}
}
return (hash & 0x7FFFFFFF);
}

posted @ 2009-04-16 22:12 杨磊阅读(554) | 评论 (0) | 编辑收藏

MySQL无法远程连接问题

今天在Lab想连一下MySQL的数据库，因为电脑不知道什么时候被人关掉了

今天又被我开起来了，之前连的都好好的，我也没做过什么特殊的处理

大概就是刚刚装好的时候，security的那个check一直过不了，然后我就搞来搞去弄了什么权限的东西

我自己都记不得是什么指令了，但是其实还是我对数据库这东西不熟悉，只懂一点点的简单的SQL

然后今天就发现连不上了，Error Number 1130，于是我去Google一下，发现这个是远程root用户没有权限的问题

原来root用户只有本地的权限，需要手动将远程的权限打开，尝试了好几种方法，最后还是下面这种方法管用

在安装mysql的机器上运行：
1、d:"mysql"bin">mysql -h localhost -u root
//这样应该可以进入MySQL服务器
2、mysql>GRANT ALL PRIVILEGES ON *.* TO 'root'@'%' WITH GRANT OPTION
//赋予任何主机访问数据的权限
3、mysql>FLUSH PRIVILEGES
//修改生效
4、mysql>EXIT
//退出MySQL服务器

想懂很多东西。

posted @ 2009-04-16 19:28 杨磊阅读(1397) | 评论 (0) | 编辑收藏

再一次见识了Java的慢……

Fence Rails

一道感觉很不错的搜索+剪枝的题目，当然剪枝策略我是参考来的，我自己想到的都是小剪枝

然后我自己用Java实现了，结果发现……，这速度啊，真让我郁闷，居然在第四个点就超时了

但是同样的算法用C++实现的代码，快了100+倍。

难怪现在还有这么多人在搞C++，也难怪有人批评说Java速度慢，还有些人在反驳

事实却是如此啊。

代码贴一下吧，TLE的，如果有人愿意的话，可以转成C++用就行了

其实我是想能不能再改进改进，让它用Java也能过，但是感觉很难，我已经想不出更好的剪枝的办法了

或者另外一个方法就是空间换时间，但是感觉已经没什么好换的了，该开的数组都开了。

1 import java.io.*;
2 import java.util.*;
3 /*
4 ID: yanglei4
5 LANG: JAVA
6 TASK:fence8
7 */
8 public class fence8{
9     int wastemax = 0;
10     int ans;
11     int N,R;
12     int[] remain;
13     int[] rails;
14     int[] from = new int[1100];
15     PrintWriter out;
16
17     public void dfs(int k, int waste) {
18         if (k < 0) {
19             out.println(ans + 1);
20             out.close();
21             System.exit(0);
22         }
23         int i;
24         if(k != ans && rails[k] == rails[k + 1]) i = from[k + 1];
25         else i = 0;
26
27         for (i = 0; i < N; i++) {
28             if (remain[i] >= rails[k]) {
29
30                 from[k] = i;
31                 remain[i] -= rails[k];
32                 if (remain[i] < rails[0]) waste += remain[i];
33                 if (waste > wastemax) {
34                     waste -= remain[i];
35                     remain[i] += rails[k];
36                     continue;
37                 }
38                 dfs(k - 1, waste);
39                 remain[i] += rails[k];
40             }
41         }
42     }
43
44     public void done() throws IOException {
45         BufferedReader f = new BufferedReader(new FileReader("fence8.in"));
46         out = new PrintWriter(new BufferedWriter(new FileWriter("fence8.out")));
47         //Read the boards
48         N = Integer.parseInt(f.readLine());
49         int[] boards = new int[N];
50         int boardSum = 0;
51         for (int i = 0; i < N; i++)
52             {
53                 boards[i] = Integer.parseInt(f.readLine());
54                 boardSum += boards[i];
55             }
56         Arrays.sort(boards);
57
58         remain = new int[N];
59         for (int i = N - 1; i >= 0; i--) remain[i] = boards[i];
60
61         // Read the rails
62         R = Integer.parseInt(f.readLine());
63         rails = new int[R];
64         int railSum = 0;
65         for (int i = 0; i < R; i++)
66             rails[i] = Integer.parseInt(f.readLine());
67         Arrays.sort(rails);
68
69         int i;
70         for (i = 0; i < R; i++) {
71             if (railSum + rails[i] > boardSum)
72                 break;
73             railSum += rails[i];
74         }
75         int k = i - 1;
76         //Try every possible number
77         for (i = k; k >= 0; i--) {
78             ans = i;
79             wastemax = boardSum - railSum;
80             railSum -= rails[i];
81             dfs(i,0);
82         }
83         out.println(0);
84         out.close();
85     }
86     public static void main(String[] args) throws IOException {
87         fence8 t = new fence8();
88         t.done();
89         System.exit(0);
90     }
91 }
92

posted @ 2009-04-16 16:50 杨磊阅读(181) | 评论 (0) | 编辑收藏

UML Structure zz from Wiki

posted @ 2009-04-15 00:10 杨磊阅读(142) | 评论 (0) | 编辑收藏

不容易啊，做了好几个小时

摘要: Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ --> 1 import java.io.*; 2 import java.util.*; &n... 阅读全文

posted @ 2009-04-14 00:53 杨磊阅读(133) | 评论 (0) | 编辑收藏

DP啊，55555～

妈了个头！！

posted @ 2009-04-12 00:42 杨磊阅读(233) | 评论 (0) | 编辑收藏

Blogjava的代码字体真丑……

等老子学好了J2EE，俺也要自己写一个blog程序

posted @ 2009-04-10 18:04 杨磊阅读(109) | 评论 (0) | 编辑收藏

感叹一下C++和Java的速度对比

真的可以用天差地别来形容啊，基本上差一个数量级

所以现在对C++还是有点留恋的，不舍得完全放弃掉

其实能把C++这么灵活的语言运用的灵活的程序员才是真正有水平的

而且C++在组织很多数据结构上确实比Java要舒服一点

可能是因为C++毕竟是从面向过程的C扩展而来的吧

不知道我这种说话会不会被人喷……

有空还是捡一下C++吧

posted @ 2009-04-09 16:51 杨磊阅读(420) | 评论 (0) | 编辑收藏

Eulerian Tour

1 # circuit is a global array
2    find_euler_circuit
3      circuitpos = 0
4      find_circuit(node 1)
5
6 # nextnode and visited is a local array
7 # the path will be found in reverse order
8   find_circuit(node i)
9
10     if node i has no neighbors then
11       circuit(circuitpos) = node i
12       circuitpos = circuitpos + 1
13     else
14       while (node i has neighbors)
15           pick a random neighbor node j of node i
16           delete_edges (node j, node i)
17           find_circuit (node j)
18       circuit(circuitpos) = node i
19       circuitpos = circuitpos + 1
20

posted @ 2009-04-04 10:21 杨磊阅读(134) | 评论 (0) | 编辑收藏

康托展开

1 unsigned long cantor(unsigned long S)
2 {
3     long x=0,i,p,k,j;
4     bool hash[8]={false};
5     for (i=8;i>=2;i--)
6     {
7         k=S>> 3*(i-1);
8         S-=k<<3*(i-1);
9         hash[k]=true;
10         p=k;
11         for (j=0;j<=k-1;j++)
12             if (hash[j])
13                 p--;
14         x+=fac[i-1]*p; //fac存的是阶乘 fac[1] = 1, fac[2] = 2, fac[3] = 6

15 }
16 return x;
17 }

其实就是求全排列中，某一个排列的序号

比如321，对应1，2，3的全排列的第6号

上面这个是8禁止存储的，有利于位操作

posted @ 2009-04-03 14:52 杨磊阅读(161) | 评论 (0) | 编辑收藏

Bellman-Ford Algo.

http://www.csanimated.com/animation.php?t=Bellman-Ford_algorithm

1 procedure BellmanFord(list vertices, list edges, vertex source)
2    // This implementation takes in a graph, represented as lists of vertices
3    // and edges, and modifies the vertices so that their distance and
4    // predecessor attributes store the shortest paths.
5
6    // Step 1: Initialize graph
7    for each vertex v in vertices:
8        if v is source then v.distance := 0
9        else v.distance := infinity
10        v.predecessor := null
11
12    // Step 2: relax edges repeatedly
13    for i from 1 to size(vertices)-1:
14        for each edge uv in edges: // uv is the edge from u to v
15            u := uv.source
16            v := uv.destination
17            if v.distance > u.distance + uv.weight:
18                v.distance := u.distance + uv.weight
19                v.predecessor := u
20
21    // Step 3: check for negative-weight cycles
22    for each edge uv in edges:
23        u := uv.source
24        v := uv.destination
25        if v.distance > u.distance + uv.weight:
26            error "Graph contains a negative-weight cycle"
27

An implementation from wiki

1 #include <limits.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4
5 /* Let INFINITY be an integer value not likely to be
6    confused with a real weight, even a negative one. */
7 #define INFINITY ((1 << 14)-1)
8
9 typedef struct {
10     int source;
11     int dest;
12     int weight;
13 } Edge;
14
15 void BellmanFord(Edge edges[], int edgecount, int nodecount, int source)
16 {
17     int *distance = malloc(nodecount * sizeof *distance);
18     int i, j;
19
20     for (i=0; i < nodecount; ++i)
21       distance[i] = INFINITY;
22     distance[source] = 0;
23
24     for (i=0; i < nodecount; ++i) {
25        int somethingchanged = 0;
26        for (j=0; j < edgecount; ++j) {
27             if (distance[edges[j].source] != INFINITY) {
28                 int new_distance = distance[edges[j].source] + edges[j].weight;
29                 if (new_distance < distance[edges[j].dest]) {
30                   distance[edges[j].dest] = new_distance;
31                   somethingchanged = 1;
32                 }
33             }
34         }
35         /* if one iteration had no effect, further iterations will have no effect either */
36         if (!somethingchanged) break;
37     }
38
39     for (i=0; i < edgecount; ++i) {
40         if (distance[edges[i].dest] > distance[edges[i].source] + edges[i].weight) {
41             puts("Negative edge weight cycles detected!");
42             free(distance);
43             return;
44         }
45     }
46
47     for (i=0; i < nodecount; ++i) {
48         printf("The shortest distance between nodes %d and %d is %d\n",
49             source, i, distance[i]);
50     }
51
52     free(distance);
53     return;
54 }
55
56 int main(void)
57 {
58     /* This test case should produce the distances 2, 4, 7, -2, and 0. */
59     Edge edges[10] = {{0,1, 5}, {0,2, 8}, {0,3, -4}, {1,0, -2},
60                       {2,1, -3}, {2,3, 9}, {3,1, 7}, {3,4, 2},
61                       {4,0, 6}, {4,2, 7}};
62     BellmanFord(edges, 10, 5, 4);
63     return 0;
64 }

posted @ 2009-04-03 14:09 杨磊阅读(245) | 评论 (0) | 编辑收藏

QSort Algo.

  1 /*************************************************************************
  2  *  Compilation:  javac QuickSort.java
  3  *  Execution:    java QuickSort N
  4  *
  5  *  Generate N random real numbers between 0 and 1 and quicksort them.
  6  *
  7  *  On average, this quicksort algorithm runs in time proportional to
  8  *  N log N, independent of the input distribution. The algorithm
  9  *  guards against the worst-case by randomly shuffling the elements
10  *  before sorting. In addition, it uses Sedgewick's partitioning
11  *  method which stops on equal keys. This protects against cases
12  *  that make many textbook implementations, even randomized ones,
13  *  go quadratic (e.g., all keys are the same).
14  *
15  *************************************************************************/
16
17 public class QuickSort {
18     private static long comparisons = 0;
19     private static long exchanges   = 0;
20
21    /***********************************************************************
22     *  Quicksort code from Sedgewick 7.1, 7.2.
23     ***********************************************************************/
24     public static void quicksort(double[] a) {
25         shuffle(a);                        // to guard against worst-case
26         quicksort(a, 0, a.length - 1);
27     }
28
29     // quicksort a[left] to a[right]
30     public static void quicksort(double[] a, int left, int right) {
31         if (right <= left) return;
32         int i = partition(a, left, right);
33         quicksort(a, left, i-1);
34         quicksort(a, i+1, right);
35     }
36
37     // partition a[left] to a[right], assumes left < right
38     private static int partition(double[] a, int left, int right) {
39         int i = left - 1;
40         int j = right;
41         while (true) {
42             while (less(a[++i], a[right]))      // find item on left to swap
43                 ;                               // a[right] acts as sentinel
44             while (less(a[right], a[--j]))      // find item on right to swap
45                 if (j == left) break;           // don't go out-of-bounds
46             if (i >= j) break;                  // check if pointers cross
47             exch(a, i, j);                      // swap two elements into place
48         }
49         exch(a, i, right);                      // swap with partition element
50         return i;
51     }
52
53     // is x < y ?
54     private static boolean less(double x, double y) {
55         comparisons++;
56         return (x < y);
57     }
58
59     // exchange a[i] and a[j]
60     private static void exch(double[] a, int i, int j) {
61         exchanges++;
62         double swap = a[i];
63         a[i] = a[j];
64         a[j] = swap;
65     }
66
67     // shuffle the array a[]
68     private static void shuffle(double[] a) {
69         int N = a.length;
70         for (int i = 0; i < N; i++) {
71             int r = i + (int) (Math.random() * (N-i));   // between i and N-1
72             exch(a, i, r);
73         }
74     }
75
76
77
78     // test client
79     public static void main(String[] args) {
80         int N = Integer.parseInt(args[0]);
81
82         // generate N random real numbers between 0 and 1
83         long start = System.currentTimeMillis();
84         double[] a = new double[N];
85         for (int i = 0; i < N; i++)
86             a[i] = Math.random();
87         long stop = System.currentTimeMillis();
88         double elapsed = (stop - start) / 1000.0;
89         System.out.println("Generating input:  " + elapsed + " seconds");
90
91         // sort them
92         start = System.currentTimeMillis();
93         quicksort(a);
94         stop = System.currentTimeMillis();
95         elapsed = (stop - start) / 1000.0;
96         System.out.println("Quicksort:   " + elapsed + " seconds");
97
98         // print statistics
99         System.out.println("Comparisons: " + comparisons);
100         System.out.println("Exchanges:   " + exchanges);
101     }
102 }
103

posted @ 2009-04-01 15:49 杨磊阅读(149) | 评论 (0) | 编辑收藏

Prim Algo.

  # distance(j) is distance from tree to node j
  # source(j) is which node of so-far connected MST
  #                      is closest to node j
1   For all nodes i
2     distance(i) = infinity        # no connections
3     intree(i) = False             # no nodes in tree
4     source(i) = nil

5   treesize = 1                    # add node 1 to tree
6   treecost = 0
7   intree(1) = True
8   For all neighbors j of node 1   # update distances
9      distance(j) = weight(1,j)
10     source(j) = 1

11   while (treesize < graphsize)
12     find node with minimum distance to tree; call it node i
13     assert (distance(i) != infinity, "Graph Is Not Connected")

    # add edge source(i),i to MST
14     treesize = treesize + 1
15     treecost = treecost + distance(i)
16     intree(i) = True              # mark node i as in tree

    # update distance after node i added
17     for all neighbors j of node i
18       if (distance(j) > weight(i,j))
19         distance(j) = weight(i,j)
20         source(j) = i

posted @ 2009-04-01 13:16 杨磊阅读(121) | 评论 (0) | 编辑收藏

人生悲凉莫过于写了200行代码最后发现TLE

摘要: 留个纪念吧 Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ --> 1 import java.io.*; 2 import java.util... 阅读全文

posted @ 2009-03-31 18:08 杨磊阅读(179) | 评论 (0) | 编辑收藏

Dijkstra's algorithm & The Floyd-Warshall Algorithm

# distance(j) is distance from source vertex to vertex j
# parent(j) is the vertex that precedes vertex j in any shortest path
#                  (to reconstruct the path subsequently)

1 For all nodes i
2     distance(i) = infinity          # not reachable yet
3     visited(i) = False
4     parent(i) = nil # no path to vertex yet

5 distance(source) = 0 # source -> source is start of all paths
6 parent(source) = nil
7   8 while (nodesvisited < graphsize)
9     find unvisited vertex with min distance to source; call it vertex i
10     assert (distance(i) != infinity, "Graph is not connected")

11     visited(i) = True # mark vertex i as visited

    # update distances of neighbors of i
12     For all neighbors j of vertex i
13         if distance(i) + weight(i,j) < distance(j) then
14             distance(j) = distance(i) + weight(i,j)
15             parent(j) = i

# dist(i,j) is "best" distance so far from vertex i to vertex j

# Start with all single edge paths.
For i = 1 to n do
    For j = 1 to n do
        dist(i,j) = weight(i,j)

For k = 1 to n do # k is the `intermediate' vertex
    For i = 1 to n do
        For j = 1 to n do
            if (dist(i,k) + dist(k,j) < dist(i,j)) then # shorter path?
                dist(i,j) = dist(i,k) + dist(k,j)

posted @ 2009-03-30 20:44 杨磊阅读(243) | 评论 (0) | 编辑收藏

DP啊，为啥我还是不会呢……

高中的时候对DP就没有感觉，到现在还是不行，尽管对于递推这种东西已经比原来有感觉的多了

但是还是没办法弄出来状态划分啊什么的，难道真的需要多做题积累经验么

一直觉得DP是很神奇的东西，因为感觉很玄乎，不像其他的算法那样，基本都有固定套路的

DP相对来说很灵活，也正是如此所以难以掌握吧

没办法啦，还是多多努力吧，多做题，多见识一下，大概就能掌握了

熟能生巧么，谁让我天生没长懂DP的脑袋呢

Dev+Algo两不误，我的追求

posted @ 2009-03-29 22:55 杨磊阅读(122) | 评论 (0) | 编辑收藏

Unit Testing Tips from TopCoder

ALWAYS implement a tested version of the demonstration from the component specification.
ALWAYS provide a meaningful message in assert and fail calls.
ALWAYS document your test code as well as you document your component code.
Break up your tests into discrete TestCase classes. If one TestCase becomes unmanageable, don't hesitate to break it into two or more classes.
Break up your tests within those classes into the smallest functions possible; this way, it is clear which areas of the component are failing. You can then use the number of tests passing and failing as a completion metric, as well.
Reduce code duplication and increase robustness with setUp() and tearDown().
Test every public function for as much valid and invalid input as time allows.
Test expected component processes: loading, processing data, and saving data, for instance.
Don't forget to clean up your environment! Unit tests should leave the system in the same state they found it in; there should be no persistent changes. This is checked during development review.
Test classes are normal classes in every respect, except that they have special significance to the testing framework. These classes can inherit from a class intermediate between themselves and the final test. They may contain methods other than test methods. They may have state.
Because they are in the same package or namespace as the component classes they test, the unit tests can access package-private and protected classes and their members.
Interfaces cannot be tested directly, but methods that accept interface arguments can be presented with alternative implementations. This technique has great potential for verifying that the component does the expected things with and to its interface-typed fields and method arguments, and that it reacts correctly to exceptions thrown by methods invoked on such arguments.

posted @ 2009-03-28 21:38 杨磊阅读(145) | 评论 (0) | 编辑收藏

抽屉原理

对一块3行7列的长方形阵列中的小方格的每一格任意染成黑色或白色，求证：在这个长方形中，一定有一个由小方格组成的长方形，它的四个角上的小方格同色。
证法1：每一列的三个格用黑、白两种颜色染色.所有可能的染法只有如下图中的八种

如果在所染色的3行7列阵列中某一列是第（1）种方式，即三格均为白色，则其余6列中只要再有第（1）（2）（3）（4）种方式之一（即该列中至少有两个白格），那么显然存在一个四角格都是白色的长方形.若第（1）、（2）、（3）、（4）种方式均未出现，那么其余6列就只能是（5）、（6）、（7）、（8）这四种方式，根据抽屉原理，其中至少有两列染色方式完全一样.又（5）～（8）中每一列至少有两格染黑色，所以一定存在一个长方形，它的四角格颜色都是黑色。
同理可知，如果有一列是第（8）种方式，即三格均为黑色，那么也存在四角同色的长方形。
如果在7列中（1）、（8）两种方式都未出现，则只有（2）、（3）、（4）、（5）、（6）、（7）这六种方式染这7列，根据抽屉原理，至少有两列染色方式完全一样，所以仍然存在四角同色的长方形。
证法2：第一行有7个小方格，用黑白两种颜色去染，根据抽屉原理，至少有四个方格所染颜色相同，不妨设第一行有4个黑方格.再看第二行，如果在第一行的四个黑方格下面的四格中有两格是黑色，则结论显然成立.否则第二行这四个格中至少有3个白色方格。
再看第三行.根据抽屉原理，在第三行的位于第二行的3个白格下面的3个格中必至少有两格同色.如果有两格为白色，则与第二行构成四角白色的长方形；如果没有两格白色，那么必有两格为黑色，则与第一行构成四角黑色的长方形。

posted @ 2009-03-03 14:34 杨磊阅读(288) | 评论 (0) | 编辑收藏

今天开始看The Art of Computer Programming

att

posted @ 2009-02-28 00:08 杨磊阅读(164) | 评论 (0) | 编辑收藏

二维图最大匹配--匈牙利算法

bool find(int a)
{
        for(int i=1;i<=m;i++)
        {
                if(g[a][ i ]==1&&!b[ i ])
                {
                        b[ i ]=true;
                        if(link[ i ]==0||find(link[ i ]))
                        {
                                link[ i ]=a;
                                return true;
                        }
                }
        }
        return false;
}

int main()
{
        while(init())
        {
                for(int i=1;i<=n;i++)
                {
                        memset(b,0,sizeof(b));
                        if(find(i))ans++;
                }
                printf("%d\n",ans);
        }
}

posted @ 2009-02-25 14:42 杨磊阅读(162) | 评论 (0) | 编辑收藏

Problems Left To Do

SRM144 DIV1 1100	图形学

posted @ 2009-02-20 23:50 杨磊阅读(102) | 评论 (0) | 编辑收藏

Games Wanted To Play

X	Titan Quest
V	Warhammer 40,000 Dawn of War II
	StarCraft II
	Diablo III

Titan Quest 实在是没心思玩，这个ARPG的游戏，居然没有大范围杀伤的技能，或者是我没找到

而且就算有，比如战斗专精的战风，也是伤害低的可怜，根本不想Diablo里面法师的魔法那么强大

差不多是完全靠剧情来打的游戏，屏幕上的怪，无论大的小的，我都只能一个一个的打，太费神了，还是算了

Warhammer 40,000 Dawn of War II，真是好游戏啊！可惜电脑不行，否则效果全开的话，一定是非常非常震撼的

也许等过个几年，我换电脑了，到时候再来尝试一下效果全开是什么感觉

To Be Continued....

有朝一日希望我有机会能玩上这些游戏，但我估计是梦想了

“游戏你的游戏，不要被游戏所游戏”

posted @ 2009-02-16 18:41 杨磊阅读(101) | 评论 (0) | 编辑收藏

Endianness from wikipedia

http://en.wikipedia.org/wiki/Endianness

posted @ 2009-02-14 00:27 杨磊阅读(166) | 评论 (0) | 编辑收藏

Big and Little Endian zz from UMD

Big and Little Endian

Basic Memory Concepts

In order to understand the concept of big and little endian, you need to understand memory. Fortunately, we only need a very high level abstraction for memory. You don't need to know all the little details of how memory works.

All you need to know about memory is that it's one large array. But one large array containing what? The array contains bytes. In computer organization, people don't use the term "index" to refer to the array locations. Instead, we use the term "address". "address" and "index" mean the same, so if you're getting confused, just think of "address" as "index".

Each address stores one element of the memory "array". Each element is typically one byte. There are some memory configurations where each address stores something besides a byte. For example, you might store a nybble or a bit. However, those are exceedingly rare, so for now, we make the broad assumption that all memory addresses store bytes.

I will sometimes say that memory is byte-addresseable. This is just a fancy way of saying that each address stores one byte. If I say memory is nybble-addressable, that means each memory address stores one nybble.

Storing Words in Memory

We've defined a word to mean 32 bits. This is the same as 4 bytes. Integers, single-precision floating point numbers, and MIPS instructions are all 32 bits long. How can we store these values into memory? After all, each memory address can store a single byte, not 4 bytes.

The answer is simple. We split the 32 bit quantity into 4 bytes. For example, suppose we have a 32 bit quantity, written as 90AB12CD₁₆, which is hexadecimal. Since each hex digit is 4 bits, we need 8 hex digits to represent the 32 bit value.

So, the 4 bytes are: 90, AB, 12, CD where each byte requires 2 hex digits.

It turns out there are two ways to store this in memory.

Big Endian

In big endian, you store the most significant byte in the smallest address. Here's how it would look:

Address	Value
1000	90
1001	AB
1002	12
1003	CD

Little Endian

In little endian, you store the least significant byte in the smallest address. Here's how it would look:

Address	Value
1000	CD
1001	12
1002	AB
1003	90

Notice that this is in the reverse order compared to big endian. To remember which is which, recall whether the least significant byte is stored first (thus, little endian) or the most significant byte is stored first (thus, big endian).

Notice I used "byte" instead of "bit" in least significant bit. I sometimes abbreciated this as LSB and MSB, with the 'B' capitalized to refer to byte and use the lowercase 'b' to represent bit. I only refer to most and least significant byte when it comes to endianness.

Which Way Makes Sense?

Different ISAs use different endianness. While one way may seem more natural to you (most people think big-endian is more natural), there is justification for either one.

For example, DEC and IBMs(?) are little endian, while Motorolas and Suns are big endian. MIPS processors allowed you to select a configuration where it would be big or little endian.

Why is endianness so important? Suppose you are storing int values to a file, then you send the file to a machine which uses the opposite endianness and read in the value. You'll run into problems because of endianness. You'll read in reversed values that won't make sense.

Endianness is also a big issue when sending numbers over the network. Again, if you send a value from a machine of one endianness to a machine of the opposite endianness, you'll have problems. This is even worse over the network, because you might not be able to determine the endianness of the machine that sent you the data.

The solution is to send 4 byte quantities using network byte order which is arbitrarily picked to be one of the endianness (not sure if it's big or little, but it's one of them). If your machine has the same endianness as network byte order, then great, no change is needed. If not, then you must reverse the bytes.

History of Endian-ness

Where does this term "endian" come from? Jonathan Swift was a satirist (he poked fun at society through his writings). His most famous book is "Gulliver's Travels", and he talks about how certain people prefer to eat their hard boiled eggs from the little end first (thus, little endian), while others prefer to eat from the big end (thus, big endians) and how this lead to various wars.

Of course, the point was to say that it was a silly thing to debate over, and yet, people argue over such trivialities all the time (for example, should braces line in parallel or not? vi or emacs? UNIX or Windows).

Misconceptions

Endianness only makes sense when you want to break a large value (such as a word) into several small ones. You must decide on an order to place it in memory.

However, if you have a 32 bit register storing a 32 bit value, it makes no sense to talk about endianness. The register is neither big endian nor little endian. It's just a register holding a 32 bit value. The rightmost bit is the least significant bit, and the leftmost bit is the most significant bit.

There's no reason to rearrange the bytes in a register in some other way.

Endianness only makes sense when you are breaking up a multi-byte quantity, and attempting to store the bytes at consecutive memory locations. In a register, it doesn't make sense. A register is simply a 32 bit quantity, b₃₁....b₀, and endianness does not apply to it.

With regard to endianness, You may argue there's a very natural way to store 4 bytes in 4 consecutive addresses, and that the other way looks strange. In particular, it looks "backwards". However, what's natural to you may not be natural to someone else. The fact of the matter is that the word is split in 4 bytes, and most people would agree that you need some order to place it in memory.

C-style strings

Once you start thinking about endianness, you begin to think it applies to everything. Before you see big or little endian, you may have had no idea it even existed. That's because it's reasonably well-hidden from you.

If you do bitwise/bitshift operations on an int, you don't notice the endianness. The machine arranges the multiple bytes so the least significant byte is still the least significant byte (e.g., b_7-0) and the most significant byte is still the most significant byte (e.g., b_31-24).

So, it's natural to think whether strings might be saved in some sort of strange order, depending on the machine.

This is where it's useful to think about all the facts you know about arrays. A C-style string, after all, is still an array of characters.

Here are some facts you should know about C-style strings and arrays.

C-style strings are stored in arrays of characters.
Each character requires one byte of memory, since characters are represented in ASCII (in the future, this could change, as Unicode becomes more popular).
In an array, the address of consecutive array elements increases. Thus, & arr[ i ] is less than & arr[ i + 1 ].
What's not as obvious is that if something is stored in increasing addresses in memory, it's going to be stored in increasing "addresses" in a file. When you write to a file, you usually specify an address in memory, and the number of bytes you wish to write to the file starting at that address.

So, let's imagine some C-style string in memory. You have the word "cat". Let's pretend 'c' is stored at address 1000. Then 'a' is stored at 1001. 't' is at 1002. The null character '"0' is at 1003.

Since C-style strings are arrays of characters, they follow the rules of characters. Unlike int or long, you can easily see the individual bytes of a C-style string, one byte at a time. You use array indexing to access the bytes (i.e., characters) of a string. You can't easily index the bytes of an int or long, without playing some pointer tricks (using reinterpret cast, for example, in C++). The individual bytes of an int are more or less hidden from you.

Now imagine writing out this string to a file using some sort of write() method. You specify a pointer to 'c', and the number of bytes you wish to print (in this case 4). The write() method proceeds byte by byte in the character string and writes it to the file, starting with 'c' and working to the null character.

Given that explanation, is it clear whether endianness matters with C-style strings? Hopefully, it is clear.

As an aside, since C++ strings are objects, it may have complicated inner structures, and so it's less obvious what a C++ string would look like when print out to a file. It's well-known what a C-style string looks like (a sequence of characters ending in a null character), which is why I've been careful to call them C-style strings.

posted @ 2009-02-14 00:25 杨磊阅读(226) | 评论 (0) | 编辑收藏

2008年11月15日

Big and Little Endian

Basic Memory Concepts

Storing Words in Memory

Big Endian

Little Endian

Which Way Makes Sense?

History of Endian-ness

Misconceptions

C-style strings

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