对每个结点 BFS 一遍，找出能被它打败和可以打败它的结点的总和 s ，如果 s == n-1 ，它的排名可以确定；

C++ vector 邻接表；

# include 
     
      # include 
      
       # include 
       
        # include 
        
         # define N 105using namespace std;int n, m;char vis[N];vector 
         
           pre[N];vector 
          
            aft[N];int bfs(int u, int d){ int i, ret, cx, nx, front, rear, size; int Q[1000*N]; Q[1] = u; memset(vis, 0, sizeof(vis)); vis[u] = 1; front = 1; rear = 2; ret = 0; while (front < rear) { cx = Q[front++]; size = d>0 ? aft[cx].size():pre[cx].size(); for (i = 0; i < size; ++i) { nx = d>0 ? aft[cx][i]:pre[cx][i]; if (!vis[nx]) { Q[rear++] = nx; vis[nx] = 1; ++ret; } } } return ret;}int main(){ int i, cnt, u, v; while (~scanf("%d%d", &n, &m)) { for (i = 1; i <= n; ++i) { pre[i].clear(); aft[i].clear(); } for (i = 0; i < m; ++i) { scanf("%d%d", &u, &v); pre[v].push_back(u); aft[u].push_back(v); } cnt = 0; for (i = 1; i <= n; ++i) { if (bfs(i, 1)+bfs(i, -1) == n-1) ++cnt; } printf("%d\n", cnt); } return 0;}
          
         
        
       
      
     

//