Poj3723--Conscription(kl+****)

Conscription

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9677		Accepted: 3440

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 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

题目： 题意读懂就好做了， 国王招兵， 需要n女m男（招1人需花费10000）， 如果女孩和男孩之间有关系且关系值为d, 则招募这两人话费可以减免d。 求花费最少， 即减免最短， 用克鲁斯卡尔求最大生成森林。

#include <cstdio>
#include <algorithm>
const int N = 20000;
using namespace std;
int father[N]; 
int Q;
struct Relationship{
    int a, b, c;
}num[50001];
bool cmp(Relationship a, Relationship b){
    return a.c > b.c;
}
void init(){
    for(int i = 0; i < Q; i++)
        father[i] = i;
}
int Find(int a){
    if(a == father[a])
        return a;
    else
        return father[a] = Find(father[a]);
}
bool Mercy(int a, int b){
    int P = Find(a);
    int O = Find(b);
    if(P != O){
        father[P] = O;
        return true;
    }
    return false;
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        int n, m, d;
        scanf("%d%d%d", &n, &m, &d);
        Q = n + m;
        init();    
        for(int i = 0; i < d; i++){
            int e, f, g;
            scanf("%d%d%d", &e, &f, &g);
            num[i].a = e; num[i].b = f+n; num[i].c = g;
        }
        sort(num, num+d, cmp);
        int sum = 0;
        for(int i = 0; i < d; i++){
            if(Mercy(num[i].a, num[i].b))
                sum += num[i].c;
        }
        printf("%d\n", Q*10000-sum);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/soTired/p/5016422.html