POJ 3723 Conscription

Conscription
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 6325   Accepted: 2184

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

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
題意：征兵：有n個女生，m個男生，其中每個人需要花費10000元。不過男生和女生之間有相互作用，可以降低費用。比如編號為1的女生和編號為1的男生之間有關系d，那麼女生已經征兵結束後，男生只需10000-d即可入伍

這道題講的很玄乎，其實就是最小生成樹。我一開始還把男生女生分別對待，即每次貪心得到最小值時，分別用男生對女生的關系和女生對男生的關系來得到最小值，然後再取。采用了prime來做。這樣確實可以做，不過男生女生數量大，超內存了。

其實仔細想一想，男生女生是圖中相同類型的點，只不過只有男生和女生之間才有路連通。所以用krusical來做，每次取最小值加入。和最小生成樹一模一樣

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int n,m;
struct node
{
    int x,y,s;
}e[50010];
int f[20010];
int find(int x)
{
    return (f[x]==x? x:f[x]=find(f[x]));
}
bool cmp(node s, node v)
{
    return s.s>v.s;
}
void krusical(int p)
{
    int i,j,ans=0;
    for (i=0; i<p; i++)
    {
        int x=find(e[i].x);
        int y=find(e[i].y);
        if (x!=y)
        {
            ans+=e[i].s;
            f[x]=f[y];
        }
    }
    cout<<10000*(n+m)-ans<<endl;
}
int main ()
{
    int i,j,t,r,s;
    cin>>t;
    while(t--)
    {
        int p=0;
        cin>>n>>m;
        for (i=0; i<=n+m; i++)
            f[i]=i;
        cin>>r;
        while(r--)
        {
            scanf("%d%d%d",&e[p].x,&j,&e[p].s);
            e[p++].y=j+n;
        }
        sort(e,e+p,cmp);
        krusical(p);
    }
    return 0;
}