有点暴力的图论…
方法:
枚举每一个节点
进行正反边搜索
正反边的遍历节点总数为N即为解

#include <iostream>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

const int MAXN = 1005;
vector<int> a[MAXN];
vector<int> ra[MAXN];
int n,m;
bool bfs(int cur)
{
    int sum = 0;
    queue<int> que;
    que.push(cur);
    int book[MAXN] = {0};

    book[cur] = 1;
    while(!que.empty())
    {
        int t = que.front();
        que.pop();

        sum++;
        for(int i=0;i < a[t].size();i++)
        {
            if(!book[a[t][i]])
            que.push(a[t][i]),book[a[t][i]] = 1;
        }
    }
    if(sum == n)
    return true;

    sum = 0;
    que.push(cur);

    int bk[MAXN] = {0};

    bk[cur] = 1;
    while(!que.empty())
    {
        int t = que.front();
        que.pop();

//      sum++;
        for(int i=0;i<ra[t].size();i++)
        {
            if(!bk[ra[t][i]])
            que.push(ra[t][i]),bk[ra[t][i]] = 1;
        }
    }

    for(int i=1;i<=n;i++)
    if(book[i] || bk[i])
    sum++;

    if(sum == n)
    return true;

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

    while(m--)
    {
        int x,y;
        scanf("%d %d",&x,&y);
        a[x].push_back(y);
        ra[y].push_back(x);
    }

    int ans= 0 ;
    for(int i=1;i<=n;i++)
    {
        if(bfs(i))
        ans++;
    }
    cout<<ans;
    return 0;
}