有向图强连通分量，板子参考Acwing1174.受欢迎的牛

#include <iostream>
#include <cstring>

using namespace std;

typedef long long LL;

const int N = 1e4+10, M = 1e5+10;

int n,m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
int stk[N], top;
bool in_stk[N];
int id[N], scc_cnt, sz[N];
int dout[N];

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

//有向图强连通分量，维护每个连通块内点的数量
void tarjan(int u){
    dfn[u] = low[u] = ++timestamp;
    stk[++top] = 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], dfn[j]);
    }

    if(dfn[u] == low[u]){
        int y;
        scc_cnt++;
        do{
            y = stk[top--];
            in_stk[y] = false;
            id[y] = scc_cnt;
            sz[scc_cnt] ++;
        }while(y != u);

    }
}

int main(){
    scanf("%d%d", &n, &m);

    memset(h, -1, sizeof h);
    while(m--){
        int a,b;
        scanf("%d%d", &a, &b);
        add(a,b);
    }

    for(int i = 1; i <= n; i++)
        if(!dfn[i])
            tarjan(i);

    LL sum = 0;
    for(int i = 1; i <= scc_cnt; i++){
        int x = sz[i];
        sum += (LL)(x-1)*x/2;   //每个连通块内部两两都是便利对，求一遍组合数C(x, 2)
    }

    printf("%lld\n", sum);

    return 0;
}