其实是给自己的一篇分享打个小广告。
Tarjan算法的简要证明
https://www.acwing.com/blog/content/1819/

代码和y总的差不多

#include <bits/stdc++.h>
using namespace std;
const int N = 10010;
const int M = 50010;
int h[N], e[M], ne[M], idx;
int n, m;
stack<int> stk;
bool in_stk[N];
int id[N], timestamp;
int scc_cnt, scc_size[N];
int deg_out[N];
int dfn[N], low[N];

void add(int a, int b){
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx ++;
}

void tarjan(int u){
    low[u] = dfn[u] = ++ timestamp;
    stk.push(u);
    in_stk[u] = true;
    for(int i = h[u]; ~i; i = ne[i]){
        int j = e[i];
        if(!dfn[j]){
            tarjan(j);
            low[u] = min(low[u], low[j]);
        }
        else if(in_stk[j]){
            low[u] = min(low[u], low[j]);
        }
    }
    if(dfn[u] == low[u]){
        int y;
        ++ scc_cnt;
        do {
            y = stk.top();
            stk.pop();
            id[y] = scc_cnt;
            in_stk[y] = false;
            scc_size[scc_cnt] ++;
        } while(y != u);
    }
}

int main(){
    memset(h, -1, sizeof h);
    cin >> n >> m;
    while(m --){
        int a, b;
        cin >> a >> b;
        add(a, b);
    }
    for(int i = 1; i <= n; i ++){
        if(!dfn[i]) tarjan(i);
    }
    for(int i = 1; i <= n; i ++){
        for(int j = h[i]; ~j; j = ne[j]){
            int k = e[j];
            int a = id[i];
            int b = id[k];
            if(a != b){
                deg_out[a] ++;
            }
        }
    }
    int sum = 0;
    int zeros = 0;
    for(int i = 1; i <= scc_cnt; i ++){
        if(!deg_out[i]){
            zeros ++;
            sum += scc_size[i];
            if(zeros > 1){
                sum = 0;
                break;
            }
        }
    }
    cout << sum << endl;
    return 0;
}