题意很简单……双方给N个人（N<=6）打三国杀1V1，你知道对手每个人能克制你哪些人，我们认为两个人对打，他能克你则他胜利，否则你胜利；某人的N个人死光就输了……你需要组织阵容按照顺序对抗对手……求有没有一个顺序，无论对手顺序如何，你都能胜……
N！枚举自己的顺序，然后N！枚举对方的顺序，O（N）验证即可……
注意字典序最小……
1 #include <cstdio>
2 #include <cstring>
3 #include <string>
4 #include <map>
5 #include <algorithm>
6 #include <iostream>
7
8 using namespace std;
9
10 map<string,int> mp;
11 int n;
12 int res[10];
13 int my[10];
14 char buf[50];
15 char name[10][50];
16 string ans;
17
18 void check() {
19 int enemy[] = {0,1,2,3,4,5,6,7,8,9};
20 do {
21 int i = 0;
22 int j = 0;
23 while (i < n && j < n) {
24 if ((1 << my[i]) & res[enemy[j]]) i ++; else j++;
25 }
26 if (i == n) return;
27 } while (next_permutation(enemy,enemy + n));
28 string tmp;
29 for (int i = 0; i < n; i++) {
30 if (i) tmp += " ";
31 tmp += name[my[i]];
32 }
33 if (ans == "" || ans > tmp) ans = tmp;
34 }
35
36 int main() {
37 int nn; scanf("%d",&nn);
38 for (int ii = 1; ii <= nn; ii++) {
39 mp.clear();
40 memset(res,0,sizeof(res));
41 scanf("%d",&n);
42 for (int i = 0; i < n; i++) {
43 scanf("%s",name[i]);
44 mp[string(name[i])] = i;
45 }
46 for (int i = 0; i < n; i++) {
47 int m; scanf("%d",&m);
48 for (int j = 0; j < m; j++) {
49 scanf("%s",buf);
50 int tmp = mp[string(buf)];
51 res[i] |= 1 << tmp;
52 }
53 }
54 for (int i = 0; i < n; i++) {
55 my[i] = i;
56 }
57 ans = "";
58 do {
59 check();
60 } while (next_permutation(my,my + n));
61 printf("Case %d: ",ii);
62 if (ans == "") {
63 puts("No");
64 } else {
65 puts("Yes");
66 for (int i = 0; i < ans.length(); i++) putchar(ans[i]);
67 putchar(10);
68 }
69 }
70 return 0;
71 }