POJ 3723 Conscription 最小生成树

题目链接：

题目

Conscription

Time Limit: 1000MS

Memory Limit: 65536K

问题描述

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.

输入

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

输出

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

样例

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

output

71071

54223

题意

现在选N个男生和M个女生进入部队，如果男生u和女生v有关系，那么如果有一个已经在部队里面了，那另一个的费用只需10000-p（关系系数）。并且每个人进入部队时他只能使用和最多一个人的关系。

问最少的花费招到所有的人。

题解

如果使用的关系之间出现了环，那么就不必有至少一个人同时使用了两个关系，所以题目就转化成了求最大生成树了。

代码

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn = 21111;

struct Edge {
	int u, v, w;
	Edge(int u, int v, int w) :u(u), v(v), w(w) {}
	bool operator < (const Edge& tmp) const {
		return w > tmp.w;
	}
};

int n, m, r;
int fa[maxn];
vector<Edge> egs;

int find(int x) { return fa[x] = fa[x] == x ? x : find(fa[x]); }

void init() {
	for (int i = 0; i <= n + m; i++) fa[i] = i;
	egs.clear();
}

int main() {
	int tc;
	scanf("%d", &tc);
	while (tc--) {
		scanf("%d%d%d", &n, &m,&r);
		init();
		while (r--) {
			int u, v, w;
			scanf("%d%d%d", &u, &v, &w);
			egs.push_back(Edge(u,v+n,w));
		}
		sort(egs.begin(), egs.end());
		int cnt = 0;
		for (int i = 0; i < egs.size(); i++) {
			Edge& e = egs[i];
			int pu = find(e.u);
			int pv = find(e.v);
			if (pu != pv) {
				cnt += e.w;
				fa[pv] = pu;
			}
		}
		printf("%d\n", (n + m) * 10000 - cnt);
	}
	return 0;
}