题目链接 Description The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms to survey their hay situation. There are N (2 <= N <= 2,000) farms (numbered 1..N); Bessie starts at Farm 1. S…

[题目链接]http://poj.org/problem?id=2395 [解题思路]找最小生成树中权值最大的那条边输出,模板过的,出现了几个问题,开的数据不够大导致运行错误,第一次用模板,理解得不够深厚且模板有错误导致提交错误   #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<algorithm&…

链接:传送门 题意:求最小生成树中的权值最大边 /************************************************************************* > File Name: poj2395.cpp > Author: WArobot > Blog: http://www.cnblogs.com/WArobot/ > Created Time: 2017年06月19日 星期一 19时00分25秒 *******************…

Out of Hay Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18472   Accepted: 7318 Description The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms to survey their hay s…

Building a Space Station Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6064   Accepted: 3015 Description You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You ar…

POJ 2395 Out of Hay 本题是要求最小生成树中的最大长度, 无向边,初始化es结构体时要加倍,别忘了init(n)并查集的初始化,同时要单独标记使用过的边数, 判断ans==n-1时,即找到了最大边. #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> #include <vector&g…

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…

POJ 1979 Red and Black (红与黑) Time Limit: 1000MS    Memory Limit: 30000K Description 题目描述 There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to…

c/c++ 用普利姆(prim)算法构造最小生成树 最小生成树(Minimum Cost Spanning Tree)的概念: ​ 假设要在n个城市之间建立公路,则连通n个城市只需要n-1条线路.这时,自然会考虑,如何在最节省经费的前提下建立这个公路网络. ​ 每2个城市之间都可以设置一条公路,相应地都要付出一定的经济代价.n个城市之间,最多可以设置n(n-1)/2条线路,那么,如何在这些可能的线路中选择n-1条,以使总的耗费最少? 普利姆(prim)算法的大致思路: ​ 大致思想是:设图G顶点…

POJ 3268 Silver Cow Party (最短路径) Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads c…