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…

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…

http://poj.org/problem?id=2728 Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 18595   Accepted: 5245 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to b…

poj 1797 Heavy Transportation Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has…

题目链接:http://poj.org/problem?id=1797 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his custom…

引用别人的解释: 题意:有n个村庄,村庄在不同坐标和海拔,现在要对所有村庄供水,只要两个村庄之间有一条路即可, 建造水管距离为坐标之间的欧几里德距离(好象是叫欧几里德距离吧),费用为海拔之差 现在要求方案使得费用与距离的比值最小 很显然,这个题目是要求一棵最优比率生成树, 概念 有带权图G, 对于图中每条边e[i], 都有benifit[i](收入)和cost[i](花费), 我们要求的是一棵生成树T, 它使得 ∑(benifit[i]) / ∑(cost[i]), i∈T 最大(或最小). 这…

题面 Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the pla…

题目链接:https://vjudge.net/problem 题意: 给各位看一下题意,算法详解看下面大佬博客吧,写的很好. 参考博客:最小k度限制生成树 - chty - 博客园  https://www.cnblogs.com/chty/p/5934669.html 代码: #include<iostream> #include<cstring> #include<algorithm> #include<queue> #include<map&g…

  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 rou…

Network Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 15268   Accepted: 5987   Special Judge Description Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the c…

题意:建光纤的时候,拉一条最长的线 思路:最大生成树 将图的n个顶点看成n个孤立的连通分支,并将所有的边按权从大到小排 边权递减的顺序,如果加入边的两个端点不在同一个根节点的话加入,并且要将其连通,否则放弃 最后剩下一个连通支 解决问题的代码: #include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<…

传送门 Heavy Transportation Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 31882   Accepted: 8445 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever…

#include<stdio.h> #define MAXN 1005 #include<iostream> #include<algorithm> #define inf 10000000 using namespace std; int _m[MAXN][MAXN]; int low_cost[MAXN]; int pre[MAXN]; unsigned prime(int n); int DFS(int i,int sum,int p); bool mark[MA…

Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place…

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…

                                                                                                Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 25729   Accepted: 7143 Description David the Great has just become the king of a…

题意: 最小生成树找出来最小的边权值总和使得n个顶点都连在一起.那么这找出来的边权值中的最大权值和最小权值之差就是本题的结果 但是题目要求让这个输出的结果最小,也就是差值最小.那么这就不是最小生成树了 题解: 思路1. 让所有边这个权值从小到大排序,之后一个权值一个权值的枚举,枚举那个权值就证明我们肯定会用到这个权值,然后让这个权值和其他边的权值做差 用这个差来代替原来的权值,再去进行最小生成树 思路2. 让所有边这个权值从小到大排序,之后一个权值一个权值的枚举,枚举到的这个权值代表我们用来连接…

POJ 1852 Ants POJ 2386 Lake Counting POJ 1979 Red and Black AOJ 0118 Property Distribution AOJ 0333 Ball POJ 3009 Curling 2.0 AOJ 0558 Cheese POJ 3669 Meteor Shower AOJ 0121 Seven Puzzle POJ 2718 Smallest Difference POJ 3187 Backward Digit Sums POJ 3…

题目:http://poj.org/problem?id=3013 看似生成树,实则最短路,可以将题意转化为点权*根到此点的边权和(最短路使其最小). 代码如下: #include<iostream> #include<cstdio> #include<cstring> #include<queue> using namespace std; typedef long long ll; queue<int>q; ],ct,s[]; ]; ll d…

题目链接:http://poj.org/problem?id=2377 于是就找了一道最大生成树的AC了一下,注意不连通的情况啊,WA了一次. /* ━━━━━┒ギリギリ♂ eye! ┓┏┓┏┓┃キリキリ♂ mind! ┛┗┛┗┛┃＼○/ ┓┏┓┏┓┃ / ┛┗┛┗┛┃ノ) ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┃┃┃┃┃┃ ┻┻┻┻┻┻ */ #include <algorithm> #include <iostream>…

百度百科:瓶颈生成树 瓶颈生成树 :无向图G的一颗瓶颈生成树是这样的一颗生成树,它最大的边权值在G的所有生成树中是最小的.瓶颈生成树的值为T中最大权值边的权. 无向图的最小生成树一定是瓶颈生成树,但瓶颈生成树不一定是最小生成树.(最小瓶颈生成树==最小生成树) 命题:无向图的最小生成树一定是瓶颈生成树. 证明:可以采用反证法予以证明. 假设最小生成树不是瓶颈树,设最小生成树T的最大权边为e,则存在一棵瓶颈树Tb,其所有的边的权值小于w(e).删除T中的e,形成两棵数T', T'',用Tb中连接T…

题目链接:http://poj.org/problem?id=3522 Slim Span Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7522   Accepted: 3988 Description Given an undirected weighted graph G, you should find one of spanning trees specified as follows. The graph G…