题意:就是求最大匹配
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<vector>
#include<limits>
#include<ctime>
#include<cassert>
#include<cstdlib>
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define cl(a,b) memset(a,b,sizeof(a));
#define LL long long
#define P pair<int,int>
#define X first
#define Y second
#define pb push_back
#define fread(a) freopen(a,"r",stdin);
#define fwrite(a) freopen(a,"w",stdout);
using namespace std;
const int maxn=5005;
const int inf=999999;
vector<int> G[maxn];
int matching[maxn];
bool vis[maxn];
int num;
bool dfs(int u){
int N=G[u].size();
for(int i=0;i<N;i++){
int v=G[u][i];
if(vis[v])continue;
vis[v]=true;
if(matching[v]==-1||dfs(matching[v])){
matching[v]=u;
return true;
}
}
return false;
}
int hungar(){
int ans=0;
cl(matching,-1);
for(int i=0;i<num;i++){
cl(vis,false);
if(dfs(i))ans++;
}
return ans/2;//无向边,算了所有。除2
}
int main(){
int T;
scanf("%d",&T);
while(T--){
int n;
scanf("%d",&n);
for(int i=0;i<=n;i++)G[i].clear();
int m=n*3/2;
while(m--){
int x,y;
scanf("%d%d",&x,&y);
x--;y--;
G[x].pb(y);//这里是无向边
G[y].pb(x);
}
num=n;
int ans=hungar();
printf("%d
",ans);
}
return 0;
}