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…

Wormholes Time Limit: 2000 MS Memory Limit: 65536 KB 64-bit integer IO format: %I64d , %I64u   Java class name: Main [Submit] [Status] [Discuss] Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A w…

题意: 农夫约翰农场里发现了很多虫洞,他是个超级冒险迷,想利用虫洞回到过去,看再回来的时候能不能看到没有离开之前的自己,农场里有N块地,M条路连接着两块地,W个虫洞,连接两块地的路是双向的,而虫洞是单向的,去到虫洞之后时间会倒退T秒,如果能遇到离开之前的自己就输出YES,反之就是NO. 分析: 就是求一幅图中有没有负权环路, 可以bellman n-1次后再跑一次看看能不能更新, 能更新说明有环. 也可以spfa记录入队次数, 入队次数大于等于N说明有负权环路 #include<cstdio>…

 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…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 29971   Accepted: 10844 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 p…

题意:John的农场里field块地,path条路连接两块地,hole个虫洞,虫洞是一条单向路,不但会把你传送到目的地,而且时间会倒退Ts.我们的任务是知道会不会在从某块地出发后又回来,看到了离开之前的自己. 思路: 这题就是判断存不存在负环回路. 前M条是双向边,后面的W是单向的负边. 为了防止出现不连通,增加一个结点作为起点.起点到所有点的长度为0 #include <iostream> #include <stdio.h> #include <string.h>…

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…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 36717   Accepted: 13438 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 p…

题意:给你m条路花费时间(双向正权路径),w个虫洞返回时间(单向负权路径),问你他能不能走一圈回到原点之后,时间倒流. 思路:题意有点难看懂,我们建完边之后找一下是否存在负权回路,存在则能,反之不能.判断负权回路可以用一个cnt,这个spfa板子里有. 代码: #include<cstdio> #include<set> #include<vector> #include<cmath> #include<queue> #include<cs…

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> #…