[Bzoj1051][HAOI2006]受欢迎的牛(tarjan)

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=1051

由题意可知，被所有牛仰慕的牛之间也互相仰慕，则最后的答案一定是唯一的强连通分量，如图：

且这个强连通分量出度为0。

所以用tarjan缩环，然后在判断出度为0的是否有且仅有一个点。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = ;
 struct node {
     int e, next;
 }edge[];
 int head[maxn * ], len;
 void add(int s, int e) {
     edge[++len].e = e;
     edge[len].next = head[s];
     head[s] = len;
 }
 int color[maxn], dfn[maxn], low[maxn], vis[maxn];
 int out[maxn];
 int num[maxn];
 int dfsnum, ansnum, top;
 stack<int>q;
 void tarjan(int x) {
     dfn[x] = low[x] = ++dfsnum;
     vis[x] = ;
     q.push(x);
     for (int i = head[x]; i; i = edge[i].next) {
         int y = edge[i].e;
         if (!dfn[y]) {
             tarjan(y);
             low[x] = min(low[x], low[y]);
         }
         else if (vis[y])
             low[x] = min(low[x], dfn[y]);
     }
     if (low[x] == dfn[x]) {
         ansnum++;
         int y;
         do {
             y = q.top();
             q.pop();
             color[y] = ansnum;
             num[ansnum]++;
             vis[y] = ;
         } while (x != y);
     }
 }
 int main() {
     int n, m, x, y;
     scanf("%d%d", &n, &m);
     for (int i = ; i <= m; i++) {
         scanf("%d%d", &x, &y);
         add(x, y);
     }
     for (int i = ; i <= n; i++)
         if (!dfn[i])
             tarjan(i);
     for (int x = ; x <= n; x++) {
         for (int i = head[x]; i; i = edge[i].next) {
             int y = edge[i].e;
             if (color[x] != color[y])
                 out[color[x]]++;
         }
     }
     int ans = , f = ;
     for (int i = ; i <= ansnum; i++)
         if (out[i] == )ans = num[i], f++;
     if (f == )
         printf("%d\n", ans);
     else
         printf("0\n");
 }