[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 = 10010;
struct node {
    int e, next;
}edge[50010];
int head[maxn * 20], 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] = 1;
    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] = 0;
        } while (x != y);
    }
}
int main() {
    int n, m, x, y;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &x, &y);
        add(x, y);
    }
    for (int i = 1; i <= n; i++)
        if (!dfn[i])
            tarjan(i);
    for (int x = 1; 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 = 0, f = 0;
    for (int i = 1; i <= ansnum; i++)
        if (out[i] == 0)ans = num[i], f++;
    if (f == 1)
        printf("%d
", ans);
    else
        printf("0
");
}