rt，如果缩点之后的图的入度是强联通分量的数量 - 1的话应该也是可以的吧？但是这样写之后 WA 44pts ，求大佬解惑qwq

代码：

//  P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G

#include<bits/stdc++.h>

using namespace std;
const int N = 1e4 + 10, M = 5e4 + 10;

int n, m;

class Graph {
private:
    vector<int> G[N];

public:
    int inDegree[N];

    void addEdge(int u, int v) {
        this->G[u].push_back(v);
        this->inDegree[v]++;
    }

    vector<int> operator[] (int idx) {
        return this->G[idx];
    }
} G1, G2;

int low[N], dfn[N], color[N], idx = 0, sccCnt = 0;
stack<int> s;
set<int> scc[N];

void tarjan(int u, Graph& G) {
    low[u] = dfn[u] = ++idx;
    s.push(u);
    for (auto v : G[u]) {
        if (!dfn[v]) {
            tarjan(v, G);
            low[u] = min(low[u], low[v]);
        } else if (!color[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (low[u] == dfn[u]) {
        sccCnt++;
        while (!s.empty()) {
            int x = s.top();
            s.pop();
            color[x] = sccCnt;
            scc[sccCnt].insert(x);
            if (x == u) break;
        }
    }
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        G1.addEdge(u, v);
    }
    for (int i = 1; i <= n; i++) {
        if (!dfn[i]) tarjan(i, G1);
    }
    for (int u = 1; u <= n; u++) {
        for (auto v : G1[u]) {
            if (color[u] != color[v]) {
                G2.addEdge(color[u], color[v]);
            }
        }
    }
    long long ans = 0;
    for (int i = 1; i <= sccCnt; i++) {
        if (G2.inDegree[i] == sccCnt - 1) ans += scc[i].size();
    }
    printf("%d\n", ans);
    return 0;
}