poj 3723 Conscription

最新推荐文章于 2020-09-28 00:10:33 发布

转载最新推荐文章于 2020-09-28 00:10:33 发布 · 154 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/8023spz/p/9170423.html

本文介绍了一个基于Kruskal算法求解最小生成树的问题实例。主人公Windy希望通过招募士兵来组建军队保护国家，利用女孩与男孩之间的特定关系减少成本。文章详细描述了输入输出格式、样例及解析过程，并提供了完整的C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

最小生成树，Kruskall算法。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
struct rela {
    int x,y,d;
}s[50000];
int t,n,m,r;
int f[20000];
void init() {
    for(int i = 0;i < (n + m);i ++) {
        f[i] = i;
    }
}
int getf(int x) {
    if(x != f[x])f[x] = getf(f[x]);
    return f[x];
}
int mer(int x,int y) {
    int xx = getf(x);
    int yy = getf(y);
    if(xx == yy)return 1;
    f[xx] = yy;
    return 0;
}
bool cmp(rela a,rela b) {
    return a.d > b.d;
}
int main() {
    scanf("%d",&t);
    while(t --) {
        scanf("%d%d%d",&n,&m,&r);
        int ans = (n + m) * 10000;
        init();
        for(int i = 0;i < r;i ++) {
            scanf("%d%d%d",&s[i].x,&s[i].y,&s[i].d);
            s[i].y += n;
        }
        sort(s,s + r,cmp);
        for(int i = 0;i < r;i ++) {
            if(mer(s[i].x,s[i].y))continue;
            ans -= s[i].d;
        }
        printf("%d\n",ans);
    }
}