题目连接 http://poj.org/problem?id=2377 Bad Cowtractors Description Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found…

http://poj.org/problem?id=2377 bessie要为FJ的N个农场联网,给出M条联通的线路,每条线路需要花费C,因为意识到FJ不想付钱,所以bsssie想把工作做的很糟糕,她想要花费越多越好,并且任意两个农场都需要连通,并且不能存在环.后面两个条件保证最后的连通图是一棵树. 输出最小花费,如果没办法连通所有节点输出-1. 最大生成树问题,按边的权值从大道小排序即可,kruskal算法可以处理重边的情况,但是在处理的时候,不能仅仅因为两个节点在同一个连通子图就判断图不合法…

Bad Cowtractors Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection r…

Bad Cowtractors Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 1 Problem Description Bessie has been hired to build a cheap internet network among Farmer John's N (…

Description Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection route…

题意:给出一个图,求出其中的最大生成树= =如果无法产生树,输出-1. 思路:将边权降序再Kruskal,再检查一下是否只有一棵树即可,即根节点只有一个 #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> using namespace std; int N, M; // 节点,边的数量 struct edge { int from, to, dist;…

链接:传送门 题意:给 n 个点 , m 个关系,求这些关系的最大生成树,如果无法形成树,则输出 -1 思路:输入时将边权转化为负值就可以将此问题转化为最小生成树的问题了 /************************************************************************* > File Name: poj2377.cpp > Author: WArobot > Blog: http://www.cnblogs.com/WArobot/ &g…

传送门:http://poj.org/problem?id=2377 Bad Cowtractors Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 17373 Accepted: 7040 Description Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that…

坏的牛圈建筑 题目大意:就是现在农夫又要牛修建牛栏了,但是农夫想不给钱,于是牛就想设计一个最大的花费的牛圈给他,牛圈的修理费用主要是用在连接牛圈上 这一题很简单了,就是找最大生成树,把Kruskal算法改一下符号就好了,把边从大到小排列,然后最后再判断是否联通(只要找到他们的根节点是否相同就可以了!) #include <iostream> #include <algorithm> #include <functional> #define MAX_N 1005 #de…

#include<stdio.h> #include<stdlib.h> #define N 1100 struct node { int u,v,w; }bian[110000]; int pre[N]; int cmp(const void *a,const void *b) { return (*(struct node *)b).w-(*(struct node *)a).w; } int find(int x) {     if(x!=pre[x])         pr…