Conscription
Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24785		Accepted: 8280
Description

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

Output

For each test case output the answer in a single line.

Sample Input

2

5 5 8
4 3 6831
1 3 4583
0 0 6592
0 1 3063
3 3 4975
1 3 2049
4 2 2104
2 2 781

5 5 10
2 4 9820
3 2 6236
3 1 8864
2 4 8326
2 0 5156
2 0 1463
4 1 2439
0 4 4373
3 4 8889
2 4 3133

Sample Output

71071
54223

## 转换为最小生成树问题

AC代码：

#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;

#define MAXN 10005
typedef long long ll;
struct edge
{
int from, to, cost;
} e[5 * MAXN];

bool cmp(const edge &a, const edge &b)
{
return a.cost < b.cost;
}

int fa[3*MAXN];
int height[3*MAXN];
int n, m, r;

//初始化并查集
void init_ufs(int sz)
{
for (int i = 0; i <= sz; ++i)
fa[i] = i;

memset(height, 0, sizeof(height));
}

int find_set(const int x)
{

if (fa[x] == x)
return x;

fa[x] = find_set(fa[x]);
return fa[x];

}

bool same(const int &a, const int &b)
{
return find_set(a) == find_set(b);
}

void unite(int a, int b)
{
a = find_set(a);
b = find_set(b);
if (a != b)
{
if (height[a] < height[b])
{
fa[a] = b;
}
else
{
fa[b] = a;
if (height[a] == height[b])
++height[a];
}
}
}

int kruskal()
{
sort(e, e + r, cmp);
init_ufs(n+m);

int ans = 0;
for (int i = 0; i < r; ++i)
{
if (!same(e[i].from, e[i].to))
{
//cout<<e[i].from<<" "<<e[i].to<<endl;
unite(e[i].from, e[i].to);
ans += e[i].cost;
}
}
return ans;
}

int main()
{
int t;
scanf("%d", &t);
while (t--)
{
scanf("%d%d%d", &n, &m, &r);
int xi, yi, di;
for (int i = 0; i < r; ++i)
{
scanf("%d%d%d", &xi, &yi, &di);
//e[i] = (edge){xi, yi + n, -di};
e[i].from = xi;
e[i].to = yi+n;
e[i].cost = -di;
}
printf("%lld\n", 10000ll * (m + n) + kruskal());
}
}