田里的虫洞 题目大意:就是这个农夫的田里有一些虫洞,田有很多个点,点与点之间会存在路,走过路需要时间,并且这些点存在虫洞,可以使农夫的时间退回到时间之前,问你农夫是否真的能回到时间之前? 读完题:这一题就是很明显了,就是要你找负值圈嘛!立马上Bellman_Ford算法 #include <iostream> #include <functional> #include <algorithm> #include <queue> #define MAX_N 5…

http://poj.org/problem?id=3259 Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 37356   Accepted: 13734 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very pec…

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 you entered the wormhole! Eac…

  Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 34934   Accepted: 12752 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…

题目链接:https://vjudge.net/problem/POJ-3259 思路:求有无负环,起点随意选就可以,因为目的只是找出有没有负环,有了负环就可以让时间一直回退,那么一定能回到当初,这里我选择从1号点开始. #include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include <string> #include <vecto…

//判断负环 dist初始化为正无穷 //正环 负无穷 #include<iostream> #include<cstring> #include<queue> #include<algorithm> using namespace std; , M = ; int dist[N]; int h[N], e[M], w[M], ne[M], idx; int n,m,z; void add(int a,int b,int c) { e[idx]=b; w[i…

 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…

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> #…

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. /* ********************…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24249   Accepted: 8652 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 pa…