http://acm.hdu.edu.cn/showproblem.php?pid=1151

题意：给你一个DAG（有向无环图），要求最小顶点覆盖

题解：二分图匹配 匈牙利算法

题解：根据公式DAG最小顶点覆盖=V-二分图最大匹配

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=1000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int mp[N][N];
int col[N];
bool vis[N];
char str;
struct node{};
bool dfs(int u){
    for(int i=1;i<=n;i++){
        if(!vis[i]&&mp[u][i]){
            vis[i]=1;
            if(col[i]==-1||dfs(col[i])){
                col[i]=u;
                return 1;
            }
        }
    }
    return 0;
}
int maxmatch(){
    int res=0;
    for(int i=1;i<=n;i++){
        memset(vis,0,sizeof(vis));
        if(dfs(i))res++;
    }
    return res;
}
void init(){
    memset(mp,0,sizeof(mp));
    memset(col,-1,sizeof(col));
    ans=0;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<=m;i++){
            scanf("%d%d",&u,&v);
            mp[u][v]=1;
        }
        cout<<n-maxmatch()<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}