poj3660 floyd

 1 //Accepted    176 KB    16 ms
 2 //一头牛，如果rank是能确定的，那么能打败他的牛的个数和被他打败的牛的个数的总和为n-1
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <iostream>
 6 #include <queue>
 7 #include <cmath>
 8 #include <algorithm>
 9 using namespace std;
10 /**
11   * This is a documentation comment block
12   * 如果有一天你坚持不下去了，就想想你为什么走到这儿！
13   * @authr songt
14   */
15 const int imax_n = 105;
16 int map1[imax_n][imax_n];
17 int n,m;
18 void floyd(int map[][imax_n])
19 {
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 (map[i][k]!=0 && map[k][j]!=0)
24     {
25         map[i][j]=1;
26     }
27 }
28 int main()
29 {
30     while (scanf("%d%d",&n,&m)!=EOF)
31     {
32         memset(map1,0,sizeof(map1));
33         int x,y;
34         for (int i=0;i<m;i++)
35         {
36             scanf("%d%d",&x,&y);
37             map1[x][y]=1;
38         }
39         floyd(map1);
40         int ans=0;
41         for (int i=1;i<=n;i++)
42         {
43             int temp=0;
44             for (int j=1;j<=n;j++)
45             temp+=map1[i][j]+map1[j][i];
46             if (temp==n-1) ans++;
47         }
48         printf("%d\n",ans);
49     }
50     return 0;
51 }