Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to be his soldiers. To collect a soldier without any privilege, he must pay 10000 RMB. There are some relationships between girls and boys and Windy can use these relationships to reduce his cost. If girl x and boy y have a relationship d and one of them has been collected, Windy can collect the other one with 10000-d RMB. Now given all the relationships between girls and boys, your assignment is to find the least amount of money Windy has to pay. Notice that only one relationship can be used when collecting one soldier.

Input

The first line of input is the number of test case.
The first line of each test case contains three integers, N, M and R.
Then R lines followed, each contains three integers xi, yi and di.
There is a blank line before each test case.

1 ≤ N, M ≤ 10000
0 ≤ R ≤ 50,000
0 ≤ xi < N
0 ≤ yi < M
0 < di < 10000

解题报告：我是看着白书的做的，白书中说该图中不能存在圈，然后思路就是先把（n+m）*10000给算出来，然后存的是逆向的边，注意男士兵的节点要n+男生的编号，否则会与女士兵重复，然后女士兵是从0开始的，这道题的坑点，我果断踩坑。。。在更新p数组的时候没有更新0这个点，然后就是kruskal的模板题了。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<vector>
#include<queue>
#include<set>
#define IL inline
#define x first
#define y second
typedef long long ll;
using namespace std;
const	int N=2e4+10,M=50010;
struct node{
   
	int a;
	int b;
	int w;
	bool operator <(const node&W) const
	{
   
		return w<W.w;
	}
}edge[M];
int p[N];
int find(int x)
{
   
	if(x!=p[x])
	p[x]=find(p[x]);
	return p[x];
}
int main()
{
   
	int t;
	scanf("%d",&t);
	while(t--)
	{
   
		int n,m,z;
		scanf("%d%d%d",&n,&m,&z);
		for(int i=0;i<=n+m;i++)
		p[i]=i;
		ll ans=10000*(n+m);
		for(int i=0;i<z;i++)
		{
   
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			edge[i].a=a,edge[i].b=b+n,edge[i].w=-c;
		}
		sort(edge,edge+z);
		for(int i=0;i<z;i++)
		{
   
			int a=edge[i].a,b=edge[i].b,c=edge[i].w;
			a=find(a),b=find(b);
			if(a!=b)
			{
   
				p[a]=b;
				ans=ans+c;
			}
		}
		printf("%lld\n",ans);
	 } 
    return 0;
}