[HAOI2006]受欢迎的牛

洛谷传送门

直接tarjan求scc，然后统计出度为0的缩点，如果多余1个就输出0，只有一个就输出这个缩点里的点。

——代码

 #include <cstdio>
 #include <cstring>
 #include <stack>

 using namespace std;

 int n, m, cnt, idx, ans, k;
 int next[], to[], head[], low[], dfn[], belong[], c[];
 bool ins[];
 stack <int> s;

 inline void add(int x, int y)
 {
     to[cnt] = y;
     next[cnt] = head[x];
     head[x] = cnt++;
 }

 inline void tarjan(int u)
 {
     low[u] = dfn[u] = ++idx;
     s.push(u);
     ins[u] = ;
     int i, v;
     for(i = head[u]; i != -; i = next[i])
     {
         v = to[i];
         if(!dfn[v])
         {
             tarjan(v);
             low[u] = min(low[u], low[v]);
         }
         else if(ins[v]) low[u] = min(low[u], dfn[v]);
     }
     if(low[u] == dfn[u])
     {
         cnt++;
         do
         {
             v = s.top();
             s.pop();
             ins[v] = ;
             belong[v] = cnt;
         }while(u != v);
     }
 }

 int main()
 {
     int i, j, x, y, u, v;
     memset(head, -, sizeof(head));
     scanf("%d %d", &n, &m);
     for(i = ; i <= m; i++)
     {
         scanf("%d %d", &x, &y);
         add(x, y);
     }
     cnt = ;
     for(i = ; i <= n; i++)
      if(!dfn[i])
       tarjan(i);
     for(u = ; u <= n; u++)
      for(i = head[u]; i != -; i = next[i])
      {
          v = to[i];
          if(belong[u] != belong[v]) c[belong[u]]++;
      }
     for(i = ; i <= cnt; i++)
      if(!c[i])
      {
          ans++;
          k = i;
      }
     if(ans > )    printf("");
     else
     {
         ans = ;
         for(i = ; i <= n; i++)
          if(belong[i] == k)
           ans++;
         printf("%d", ans);
     }
     return ;
 }