H

Think:
1知识点：最短路+_Floyd()算法求传递背包数量
2反思：学习知识要努力拓展创新

建议参考博客1
建议参考博客2

以下为Accepted代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1e2 + 4;int n, e[N][N];void Floyd();int main(){    int m, i, j, u, v, sum, ans;    while(~scanf("%d %d", &n, &m)){        ans = 0;        memset(e, 0, sizeof(e));        for(i = 1; i <= m; i++){            scanf("%d %d", &u, &v);            e[u][v] = 1;        }        Floyd();        for(i = 1; i <= n; i++){            sum = 0;            for(j = 1; j <= n; j++){                if(e[i][j] || e[j][i])                    sum++;            }            if(sum == n-1)                ans++;        }        printf("%d\n", ans);    }    return 0;}void Floyd(){    int i, j, k;    for(k = 1; k <= n; k++){        for(i = 1; i <= n; i++){            for(j = 1; j <= n; j++){                if(e[i][k] && e[k][j])                    e[i][j] = 1;            }        }    }}