POJ 3259 Wormholes(Bellman-Ford)

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POJ 3259 Wormholes Bellman题解

版权声明:本文作者靖心,靖空间地址:http://blog.csdn.net/kenden23/.未经本作者同意不得转载. https://blog.csdn.net/kenden23/article/details/37737817 本题就是须要检查有没有负环存在于路径中,使用Bellman Ford算法能够检查是否有负环存在. 算法非常easy,就是在Bellman Ford后面添加一个循环推断就能够了. 题目故事非常奇怪,小心读题. #include <stdio.h> #include…

ACM: POJ 3259 Wormholes - SPFA负环判定

POJ 3259 Wormholes Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way…

最短路(Bellman_Ford) POJ 3259 Wormholes

题目传送门 /* 题意:一张有双方向连通和单方向连通的图,单方向的是负权值,问是否能回到过去(权值和为负) Bellman_Ford:循环n-1次松弛操作,再判断是否存在负权回路(因为如果有会一直减下去) 注意:双方向连通要把边起点终点互换后的边加上 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #…

poj - 3259 Wormholes (bellman-ford算法求最短路）

http://poj.org/problem?id=3259 农夫john发现了一些虫洞,虫洞是一种在你到达虫洞之前把你送回目的地的一种方式,FJ的每个农场,由n块土地(编号为1-n),M 条路,和W个虫洞组成,FJ想从一块土地开始,经过若干条路和虫洞,返回到他最初开始走的地方并且时间要在他离开之前,或者恰好等于他离开的时间. 把虫洞的时间看成负边权,就是判断从起点出发是否存在负权回路.那么就可以采用bellman-ford算法,注意数组开大点. /* ********************…

POJ 3259 Wormholes（最短路径，求负环）

POJ 3259 Wormholes(最短路径,求负环) Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE…

POJ 3259 Wormholes （Bellman_ford算法）

题目链接:http://poj.org/problem?id=3259 Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 45077 Accepted: 16625 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is ver…

poj 3259 Wormholes

题目连接 http://poj.org/problem?id=3259 Wormholes Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a tim…