Conscription POJ - 3723

傳送門

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 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

白書上的例題。

徵女兵N人，男兵M人，每徵募一個人需要花費10000美元，如果徵募中有關係的人，那麼可以減少d美元，要求徵募一個人只能利用一個關係，要求徵募所有人所需要的最小費用。

把人看成是頂點，關係看成邊，問題就轉換成求無向圖中最大權森林問題。最大權問題可以通過把所有邊權取反之後用最小生成樹來求解。

注意這裏的並查集部分必須要路徑壓縮，不然會超時到懷疑人生

#include<string>
#include<math.h>
#include<stdio.h>
#include<iostream>
#include<queue>
#include<algorithm>
#include<vector>
#include<string.h>
#include<iterator>
using namespace std;
typedef unsigned long long ll;
const int maxn = 50005;
int bin[maxn];
int n,m,r;
int V,E;
struct edge{
    int u,v,cost;
}es[maxn];

bool cmp(edge e1,edge e2)
{
    return e1.cost<e2.cost;
}


int findx(int x){
    if(bin[x] == x)
        return x;
    else{//路徑壓縮
        bin[x] = findx(bin[x]);
        return bin[x];
    }
}

void merge1(int x,int y){
    int fx = findx(x);
    int fy = findx(y);
    if(fx != fy)
        bin[fx] = fy;
}

int kruskal(){
    sort(es+1,es+E+1,cmp);
    for(int i = 1; i <= V ; i++)
        bin[i] = i;
    int res = 0;
    for(int i = 1; i <= E ; i++){
        edge e = es[i];
        if(findx(e.u)!=findx(e.v)){
            merge1(e.u,e.v);
            res += e.cost;
        }
    }
    return res;
}

int main(){
    int t;
    int x,y,d;

    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&m,&r);
        V = n+m;
        E = r;
        for(int i = 1 ; i <= r; i ++){
            scanf("%d%d%d",&x,&y,&d);
            es[i].u = x+1;
            es[i].v = n+y+1;
            es[i].cost = -d;
        }
        printf("%d\n",(m+n)*10000+kruskal());
    }
    return 0;
}