Codeforces Global Round 8 E

#include<map>
#include<queue>
#include<time.h>
#include<limits.h>
#include<cmath>
#include<ostream>
#include<iterator>
#include<set>
#include<stack>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep_1(i,m,n) for(int i=m;i<=n;i++)
#define mem(st) memset(st,0,sizeof st)
int read()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')             //判断正负
        flag=1;
    else if(ch>='0'&&ch<='9')           //得到完整的数
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef pair<double,double> pdd;
const int inf = 0x3f3f3f3f;
const int N = 4e5 + 5;
int h[N],ne[N],idx,e[N];
int n, m;
//由4/7  可知：是一个 三层完全二叉树 的子节点
//那么也就是 把三层二叉树的子节点都删掉，就行
void add(int a,int b)
{
    e[idx]=b;
    ne[idx]=h[a];
    h[a]=idx++;
}
void solve()
{
    idx=0;
    cin >> n >> m;
    for (int i = 0; i <= n; i++)
    {
        h[i]=-1;
    }
    for (int i = 1; i <= m; i++)
    {
        int u, v;
        cin >> u >> v;//按拓扑序 
        add(v,u);
    }
    vector<int>dist(n+1,0);
    vector <int> ans;
    for (int i = 1; i <= n; i++)
    {
        for (int u=h[i]; u!=-1; u=ne[u])
        {
            int j=e[u];
            if(dist[i]==-1)
                continue;
            dist[i] = max(dist[j] + 1, dist[i]);
        }
        if (dist[i] == 2)//如果是2了，那么就是说明存在一个点，到这个点的距离是3
        {
            dist[i] = -1;//在第三层，就要删掉
            ans.push_back(i);
        }
    }
    cout << ans.size() << endl;
    for (auto it : ans)
        cout << it << ' ';
    cout << endl;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
        solve();
    return 0;
}