esc uil reduce test case ive 遍歷 use cas ret

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16957		Accepted: 5896

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-dRMB. 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 x_i, y_i and d_i.
There is a blank line before each test case.

1 ≤ N, M ≤ 10000
0 ≤ R ≤ 50,000
0 ≤ x_i < N
0 ≤ y_i < M
0 < d_i < 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

Source

POJ Monthly Contest – 2009.04.05, windy7926778 大概題意就是征男女兵，然後一個兵就需要10000，而對應的男女兵有個親密度的東西，當兩者都在時候就可以減去這個值，減少花費，求最小花費，而且對應的關系只可以用一次，這就可以用來就最大生成樹來求，只遍歷一遍。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAX 50005
using namespace std;
struct mm{
int m;
int w;
int love;
}ma[MAX];
int n,m,r,f[MAX];
bool cmp(mm a,mm b){
return a.love>a.love;
}
int find(int a)
{
return f[a]==a?a:find(f[a]);
}
int kursh(){

int sum=0;
for(int i=0;i<m+n;i++)
f[i]=i;sort(ma,ma+r,cmp);
for(int i=0;i<r;i++){
int x1,x2;
x1=find(ma[i].w);
x2=find(ma[i].m);
if(x1!=x2){
f[x2]=x1;
sum+=ma[i].love;
}

}
return sum;
}
int main(){
int T;
while(~scanf("%d",&T)){
while(T--){
cin>>n>>m>>r;
for(int i=0;i<r;i++){
cin>>ma[i].w>>ma[i].m>>ma[i].love;
ma[i].m+=n;

}
cout<<(n+m)*10000-kursh()<<endl;
}
}
}

Conscription（Kruskal）