题目链接: https://vjudge.net/problem/POJ-3259 题目大意: 农夫约翰在探索他的许多农场,发现了一些惊人的虫洞.虫洞是很奇特的,因为它是一个单向通道,可让你进入虫洞的前达到目的地!他的N(1≤N≤500)个农场被编号为1..N,之间有M(1≤M≤2500)条路径,W(1≤W≤200)个虫洞.FJ作为一个狂热的时间旅行的爱好者,他要做到以下几点:开始在一个区域,通过一些路径和虫洞旅行,他要回到最开时出发的那个区域出发前的时间.也许他就能遇到自己了:).为了帮助FJ…

https://vjudge.net/problem/POJ-3259 一开始理解错题意了,以为从A->B一定得走路,B->A一定得走虫洞.emmm其实回来的时候可以路和虫洞都可以走,只要最终结果满足就好. 发现了这一点,我终于愉快地把我的floyd从wa改到了tle~ 正解:用bellman-ford判断有无负圈. #include<iostream> #include<cstdio> #include<queue> #include<cstring…

POJ3259 :Wormholes 时间限制:2000MS 内存限制:65536KByte 64位IO格式:%I64d & %I64u 描述 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 destin…

题目链接:poj3259 Wormholes 题意:虫洞问题,有n个点,m条边为双向,还有w个虫洞(虫洞为单向,并且通过时间为倒流,即为负数),问你从任意某点走,能否穿越到之前. 贴个SPFA代码: #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<stack> #include<…

题目链接:http://poj.org/problem?id=3259 Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions:75598   Accepted: 28136 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very…

Wormholes POJ-3259 这题是最短路问题中判断是否存在负圈的模板题. 判断负圈的一个关键就是理解:如果在图中不存在从s可达的负圈,最短路径不会经过一个顶点两次.while循环最多执行v-1次. 这题还需要注意的是输入的w条边的权值是负值,因为这是虫洞边. package POJ; import java.util.*; public class POJ_3259 { static int f,n,m,w; public int from,to; public double rate…

spfa判断是否存在负环,path双向,wormhole单向…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 35103   Accepted: 12805 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:有一些点.一些道路和一些虫洞,道路是双向的,连接两点,花费正的时间,而虫洞是单向的,连接两点,可以使时间倒退,求是否能够回到过去. 只要明确回到过去其实就是当出现一个负环的时候,不断沿这个环走,就能够实现时间倒退了. 然后就是判负环…… spfa版: #include<stdio.h> #include<string.h> #include<queue> using namespace std; ],next[],point[],val[],size; ],t[]…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24864   Accepted: 8869 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 DescriptionWhile 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 worm…

题目连接:http://poj.org/problem?id=3259 题意:John的农场里N块地,M条路连接两块地,W个虫洞,虫洞是一条单向路,会在你离开之前把你传送到目的地,就是当你过去的时候时间会倒退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.总的来说,就是看图中有没有负权环.有的话就是可以,没有的话就是不可以了. 分析:sfa判负环,直接建图套模板即可. #include <cstdio> #include <cstring> #includ…

题意:有n个顶点,m条边,然后有w个洞,过每个洞的时间为-ti,求是否会时光倒流 分析:就是求是否存在负圈,用Bellman-Floyd判定是否存在负圈即可,注意是无向图,所以路径是双向可达的 #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <vector> #include <algorithm> #include…

题目电波   3259 Wormholes #include<iostream> #include<cstring> #include<algorithm> #include<queue> #include<stdio.h> using namespace std; #define maxn 100010 #define inf 0x3f3f3f3f struct ac{ int to,va; ac(){} ac(int a,int b){ to…

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…

题意: 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,就是是否存在负权回路.  虽然题中没有说明起点和终点 但从1开始即可  因为无论从哪个点开始 有没有负环的情况都是一样的 spfa 判断负环: #include <iostream> #inclu…

<题目链接> 题目大意: John的农场里N块地,M条路连接两块地,W个虫洞,虫洞是一条单向路,会在你离开之前把你传送到目的地,就是当你过去的时候时间会倒退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.总的来说,就是看图中有没有负权环. 解题分析:判负环模板题,下面用的是spfa算法.判负环的依据为:如果在最短路的松弛操作中,存在进入队列次数大于n的点,则说明该图存在负环. #include <cstdio> #include <cstring&g…

http://poj.org/problem?id=3259 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 BEFOR…

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! Each of FJ's fa…

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…

版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/u011775691/article/details/27612757 非常easy的bellmanford题目.这里比較具体:http://blog.csdn.net/lyy289065406/article/details/6645790 直接代码 #include <cstdio> #include <cstdlib> #include <iostream> #inc…

自己的第一道spfa,纪念一下,顺便转载一下spfa的原理.先po代码: #include <iostream> #include <queue> using namespace std; ; ; int minimum(int a, int b){ return a > b ? b : a; } int main() { int t; cin >> t; while (t--){ int n, m, w; cin >> n >> m &g…

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: 2000MS   Memory Limit: 65536K Total Submissions: 68097   Accepted: 25374 题目链接:http://poj.org/problem?id=3259 Description: While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is ve…

题目大意:给你一张图,先输入m条双向边(权值为正),再输入w条单向边(权值为负),判断是否有负环 题目思路:bellman-ford或者SPFA都行,我用的是SPFA(因为和POJ1860类似,就不加详细注释了) #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <cstring>…

题意:有向图判负环. 解题关键:spfa算法+hash判负圈. spfa判断负环:若一个点入队次数大于节点数,则存在负环.  两点间如果有最短路,那么每个结点最多经过一次,这条路不超过$n-1$条边.” 如果一个结点经过了两次,那么我们走了一个圈.如果这个圈的权为正,显然不划算:如果是负圈,那么最短路不存在:如果是零圈,去掉不影响最优值. 也就是说,每个点最多入队$n-1$次,可以想象一下,左边$n-1$个节点全部指向右边一个节点,遍历的顺序恰好与边权顺序相反. 负圈是指圈上的总和小于0 实际只…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 39078   Accepted: 14369 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…

题目链接:http://poj.org/problem?id=3259 Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 55082   Accepted: 20543 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is ver…

英语能力差!百度的题意才读懂!就是一个判断有无负环的题.SPFA即可.,注意重边情况!! #include<iostream> //判断有无负环,spfa #include<queue> #include<cstring> #include<cstdio> using namespace std; int mark[503];int a[503][503];int d[503];int num_in[503]; bool spfa(int n) { queu…

ballman_ford 是对单源点到任意点最短路的处理方法(可以含负权边). 对所有边进行n-1次循环,(n为点得个数),如果此时源点到这条边终点的距离 大于 源点到这条边起点的距离加上路得权值就进行更新. 题目链接 题意:FJ农场主有F个农场, 在每个农场内有 N个点, M条双向路, W个单向虫洞, 每条路需要消耗一定的时间, 每个虫洞可以使得自己回到几秒前, 现在问FJ农场主可不可以遇到以前的自己. 题解: ballman_ford 判断负环, 存在负环就说明可以, 不存在就不可以. #i…