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…
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…
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:http://poj.org/problem?id=3259 题意:主要就是构造图, 然后判断,是否存在负图,可以回到原点 /* 2 3 3 1 //N, M, W 1 2 2 1 3 4 2 3 1 3 1 3 //虫洞 3 2 1 //N, M, W 1 2 3 2 3 4 3 1 8 */ #include <iostream> #include <cstring> using namespace std; + + ) * + ; + ; int N, M, W; //…
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<…
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…
Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 38300   Accepted: 14095 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…
J - Wormholes Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Description In the year 2163, wormholes were discovered. A wormhole is a subspace tunnel through space and time connecting two star systems. Wormholes…
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…
Bellman-Ford 可解决带有负权边的最短路问题 解决负权边和Dijkstra相比是一个优点,Bellman-Ford的核心代码只有4行:: u[],v[],w[] 分别存一条边的顶点.权值,dis[]存从 1 源点到各个顶点的距离 ;i<=n-;i++) ;j<=m;j++) if(dis[v[j]] > dis[u[j]]+w[j]) dis[v[j]] = dis[u[j]]+w[j]; 愿过程: 循环n-1次,把每个顶点每条边都松弛: 优化方法: ①,最坏的情况就是循环了n…