Conscription

                                            Conscription

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 4   Accepted Submission(s) : 1

Problem 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

PKU

 

题很长，大致意思就是，现在要征兵，一人需要发一万，有男有女，但是如果说男兵和女兵之间有关系，就可以省去一部分钱，现在求最少需要多少钱，输入数据表示：第一个数据表示输入的数据组数，然后一行是男女个数，以及接下来有多少行输入。先算出如果一点关系都没有需要多少钱，然后算一下最大生成树，我用的是克鲁斯卡尔，排序的时候从大到小排序，然后建树。代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int pre[100010],m,n,l;
struct node 
{
	int x,y;
	int val;
}edge[100010];
int cmp(node n1,node n2)
{
	return n1.val>n2.val;
}
void itin()
{
	for(int i=0;i<100010;i++)
	pre[i]=i;
}
int find(int x)
{
	if(x==pre[x])
	return x;
	return pre[x]=find(pre[x]);
}
bool join(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
	{
		pre[fx]=fy;
		return true;
	}
	return false;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		itin();
		scanf("%d%d%d",&m,&n,&l);
		int i,j,a,b,c;
		int sum=(n+m)*10000;
		for(i=0;i<l;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			edge[i].x=a;
			edge[i].y=b+m;
			edge[i].val=c;
		}
		sort(edge,edge+l,cmp);
		for(i=0;i<l;i++)
		{
			if(join(edge[i].x,edge[i].y))
			sum-=edge[i].val;
		}
		printf("%d
",sum);
	}
	return 0;
}