既然是拓扑图，那么可以直接使用dfs来解

C++ 代码

#include<iostream>
#include<cstring>
#include<algorithm>
#include<bitset>
using namespace std;

const int N = 30010;
int head[N] , e[N] , ne[N] , idx;
int n , m;
bitset<N> f[N];

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

void dfs(int u){
    if (f[u].count() != 0) return;
    f[u][u] = 1;
    for(int i = head[u] ; i != -1 ; i = ne[i]){
        int j = e[i];
        dfs(j);
        f[u] = f[u] | f[j];
    }
}

int main(){
    memset(head , -1 , sizeof head);
    cin >> n >> m;

    for(int i = 1 ; i <= m ; i ++){
        int a , b;
        cin >> a >> b;
        add(a , b);
    }

    for(int i = 1 ; i <= n ; i ++){
        if (f[i].count() == 0){
            dfs(i);
        }
    }

    for(int i = 1 ; i <= n ; i ++){
        cout << f[i].count() << endl;
    }
    return 0;
}