原题传送门 先用tarjan缩点形成一个DAG 然后两种做法 一种缩好点后统计每个点的入度，若入度为总数-1，就是受欢迎的牛，然后我就wa掉了，因为新边可能重复 所以换个角度，如果某个新点出度为0，那么是受欢迎的牛，但是若出度为0的点有多个，答案就是0（yy一下）
Code：
#include <bits/stdc++.h> #define maxn 100010 using namespace std; struct Edge{ int to, next; }edge[maxn]; int num, head[maxn], Index, dfn[maxn], low[maxn], vis[maxn], top, sta[maxn]; int tot, color[maxn], sum[maxn], x[maxn], y[maxn], n, m, out[maxn]; inline int read(){ int s = 0, w = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') w = -1; for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48); return s * w; } void addedge(int x, int y){ edge[++num] = (Edge){ y, head[x] }; head[x] = num; } void tarjan(int u){ dfn[u] = low[u] = ++Index; vis[u] = 1, sta[++top] = u; for (int i = head[u]; i; i = edge[i].next){ int v = edge[i].to; if (!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]); else if (vis[v]) low[u] = min(low[u], dfn[v]); } if (dfn[u] == low[u]){ ++tot; while (sta[top + 1] != u) color[sta[top]] = tot, vis[sta[top--]] = 0, ++sum[tot]; } } int main(){ n = read(), m = read(); for (int i = 1; i <= m; ++i){ x[i] = read(), y[i] = read(); addedge(x[i], y[i]); } for (int i = 1; i <= n; ++i) if (!dfn[i]) tarjan(i); num = 0; memset(head, 0, sizeof(head)); for (int i = 1; i <= m; ++i) if (color[x[i]] != color[y[i]]) addedge(color[x[i]], color[y[i]]), ++out[color[x[i]]]; int ans = 0, cnt = 0; for (int i = 1; i <= tot; ++i) if (!out[i]) ans += sum[i], ++cnt; printf("%d\n", cnt == 1 ? ans : 0); return 0; }