題目鏈接

 1 /*
 2     Name:nyoj-211-Cow Contest
 3     Copyright:
 4     Author:
 5     Date: 2018/4/27 21:02:06
 6     Description:
 7     floyd算法
 8     大佬的驚奇思路 
 9 */
10 #include <iostream>
11 #include <cstdio>
12 #include <cstring>
13 using namespace std;
14 
15 const int MAXN = 105
 
;
16 const int INF = 0x3f3f3f3f;
17 int N, g[MAXN][MAXN], M;
18 
19 void floyd() {
20     for (int k=1; k<=N; k++) {
21         for (int i=1; i<=N; i++) {
22             for (int j=1; j<=N; j++) {
23                 if (g[i][k] && g[k][j])
24                     g[i][j] = 1;//有關系 
 

25             }
26         }
27     }
28 }
29 int main()
30 {
31     while (cin>>N>>M, N+M) {
32         memset(g, 0, sizeof(g));
33         for (int i=0; i<M; i++) {
34             int x, y;
35             cin>>x>>y;
36             g[x][y] = 1;
37         }
38
 
         floyd();
39         int i, j, ans=0;
40         for (i=1; i<=N; i++) {
41             for (j=1; j<=N; j++) {
42                 if (i==j) continue;
43                 if (g[i][j] == 0 && g[j][i] == 0) break;//和其他牛中的一頭沒有關系就不能確定排名 
44             }
45             if (j > N) ans++;
46         }
47         cout<<ans<<endl;
48     }
49     return 0;
50 }

nyoj-211-Cow Contest(floyd算法)