POJ 3723 Conscription

最新推荐文章于 2018-10-30 12:52:46 发布

shyoldman

最新推荐文章于 2018-10-30 12:52:46 发布

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本文介绍了一个关于征兵的问题，通过使用Kruskal算法来解决最小生成树问题，以达到降低征兵成本的目的。该算法适用于解决包含特定男女关系及成本的场景。

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Conscription

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14103		Accepted: 4898

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

Sample Output

71071
54223

题目大意：大佬要挑n个女生和m个男生到他的军队里面，每挑到一个人，就要给人家10000RMB，如果女孩x与男孩y有关系d且他们其中一个被招了，那么招另外一个就只用支付10000-d RMB了，算出最少的费用

最小生成树，用的是Kruskal算法

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,r;
int tree[20010];
struct Node{
	int u,v,w;
}map[50010];
bool cmp(Node a,Node b){
	return a.w>b.w;
}
int find(int x){
	return x==tree[x]?x:tree[x]=find(tree[x]);
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		cin>>n>>m>>r;
		for(int i=0;i<=n+m;i++) tree[i]=i;
		int kount=0;
		while(r--){
			int x,y,z;
			scanf("%d%d%d",&x,&y,&z);
			map[kount].u=x;
			map[kount].v=y+n;
			map[kount++].w=z;
		}
		sort(map,map+kount,cmp);
		int sum=(n+m)*10000;
		for(int i=0;i<kount;i++){
			int a=find(map[i].u);
			int b=find(map[i].v);
			if(a!=b){
				sum-=map[i].w;
				tree[a]=b;
			}
		}
		printf("%d\n",sum);
	}
	return 0;
}