zoukankan      html  css  js  c++  java
  • POJ 3723 Conscription 最小生成树

    题目链接:

    题目

    Conscription
    Time Limit: 1000MS
    Memory Limit: 65536K

    问题描述

    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.

    输入

    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

    输出

    For each test case output the answer in a single line.

    样例

    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

    output
    71071
    54223

    题意

    现在选N个男生和M个女生进入部队,如果男生u和女生v有关系,那么如果有一个已经在部队里面了,那另一个的费用只需10000-p(关系系数)。并且每个人进入部队时他只能使用和最多一个人的关系。
    问最少的花费招到所有的人。

    题解

    如果使用的关系之间出现了环,那么就不必有至少一个人同时使用了两个关系,所以题目就转化成了求最大生成树了。

    代码

    #include<iostream>
    #include<cstdio>
    #include<vector>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    
    const int maxn = 21111;
    
    struct Edge {
    	int u, v, w;
    	Edge(int u, int v, int w) :u(u), v(v), w(w) {}
    	bool operator < (const Edge& tmp) const {
    		return w > tmp.w;
    	}
    };
    
    int n, m, r;
    int fa[maxn];
    vector<Edge> egs;
    
    int find(int x) { return fa[x] = fa[x] == x ? x : find(fa[x]); }
    
    void init() {
    	for (int i = 0; i <= n + m; i++) fa[i] = i;
    	egs.clear();
    }
    
    int main() {
    	int tc;
    	scanf("%d", &tc);
    	while (tc--) {
    		scanf("%d%d%d", &n, &m,&r);
    		init();
    		while (r--) {
    			int u, v, w;
    			scanf("%d%d%d", &u, &v, &w);
    			egs.push_back(Edge(u,v+n,w));
    		}
    		sort(egs.begin(), egs.end());
    		int cnt = 0;
    		for (int i = 0; i < egs.size(); i++) {
    			Edge& e = egs[i];
    			int pu = find(e.u);
    			int pv = find(e.v);
    			if (pu != pv) {
    				cnt += e.w;
    				fa[pv] = pu;
    			}
    		}
    		printf("%d
    ", (n + m) * 10000 - cnt);
    	}
    	return 0;
    }
  • 相关阅读:
    C#与数据库访问技术总结(十四)之DataAdapter对象
    《运维三十六计》:运维生存宝典
    企业运维几百个重点面试题汇总(老男孩)
    5、KVM虚拟化典型案例:生产环境问题案例与分析
    Linux运维必会的实战编程笔试题(19题)
    面试中常见的 MySQL 考察难点和热点
    Linux系统运维常见面试简答题(36题)
    一键备份脚本 backup.sh
    Linux常用的200个命令总结分类
    2、KVM虚拟化CPU、内存、网络、磁盘技术及性能调优方法
  • 原文地址:https://www.cnblogs.com/fenice/p/5641222.html
Copyright © 2011-2022 走看看