poj 3723 Conscription(最小生成树)

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#最小生成树 #kruskal

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本文介绍了一种通过算法确定征兵最优成本的方法。考虑到男女兵之间的特定关系，该算法能够计算出征召最少成本。利用Kruskal算法解决此问题，并通过调整边的权重将最大权森林问题转化为最小生成树问题。

Conscription

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11677		Accepted: 4124

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

Sample Output

71071
54223

Source

POJ Monthly Contest – 2009.04.05, windy7926778

男女都是从0开始，为了在一颗树上，可以在男兵的序号加上n，即从0-n, 0-m变成了0-m+n。每条边的权值是可以省去的费用，如过没有边，总费用是(m+n)*10000，现在可以将边的权值改成负值，这样就把最大权森林问题变成了最小生成树问题了。

Kruskal算法实现代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1e5+5;
int n, m, r, k, pre[maxn];

struct node
{
    int u, v, w;
    node() {}
    node(int uu, int vv, int ww): u(uu), v(vv), w(ww) {}
    bool operator < (const node &a) const
    {
        return w < a.w;
    }
}edge[maxn];

int Find(int x)
{
    int r = x;
    while(pre[r] != r) r = pre[r];
    int i = x, j;
    while(i != r)
    {
        j = pre[i];
        pre[i] = r;
        i = j;
    }
    return r;
}

bool join(int x, int y)
{
    int a = Find(x);
    int b = Find(y);
    if(a != b)
    {
        pre[b] = a;
        return true;
    }
    return false;
}

int main(void)
{
    int t;
    cin >> t;
    while(t--)
    {
        k = 0;
        scanf("%d%d%d", &n, &m, &r);
        for(int i = 0; i < m+n; i++) pre[i] = i;
        int R = r;
        while(R--)
        {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            v += n;
            w = -w;
            edge[k++] = node(u, v, w);
        }
        sort(edge, edge+r);
        int ans = (n+m)*10000, count = 0;
        for(int i = 0; i < r; i++)
        {
            if(join(edge[i].u, edge[i].v))
            {
                count++;
                ans += edge[i].w;
            }
            if(count == n+m-1) break;
        }
        printf("%d\n", ans);
    }
    return 0;
}