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POJ 3259 Wormholes 虫洞问题
问题的链接在这里。 很显然本题是要检测负权回路（沿着一条路径回到初始点，所需要的耗费为负值。） 首先在这里介绍一下bellman-ford算法和改进后的spfa算法。 1.bellman-ford算法 bellman-ford也是一种可以求单元最短路径的算法，与Dijkstra算法不同的是，它可以用来判断负权边是否存在。poj 3259 Wormholes spfa判负环
#第一以为是直接传送看贝西会不会无限循环那道题，直接洛谷搬过来交了一发，wa了 哦，不是，是时间旅行者JOHN. #建图+判负环 虫洞单向边，权值为负； 路径双向边，权值为正； #判负环的条件是入队次数>n --- 坑点： 听说poj的评测机菜到动用min不行要手写if； 还听说连三目运算都会被卡; 我的坑点是老poj-3259(判断负权回路)
题目：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 time that is BEFORE you entered thWormholes POJ - 3259 spfa判断负环
//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; const int N=1e5,INF=0x3f3f3f3f; int dist[N]; int h[N],e[N],ne[N],w[N],idx; int n,m,z; void add(int a,iPOJ - 3259 - Wormholes = SPFA
http://poj.org/problem?id=3259 看见500的数据量莽了一发Floyd，还T了。讲道理应该不至于的，换个快读试试。 #include<algorithm> #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<map> #include<set> #include<stack> #POJ 3259 Wormholes
题目链接：https://vjudge.net/problem/POJ-3259 题目大意 略。 分析 Bellman-Ford 判断负环模板题。 代码如下 1 #include <cmath> 2 #include <ctime> 3 #include <iostream> 4 #include <string> 5 #include <vector> 6 #include <cstdio> 7 #ipoj-3259 Wormholes（无向、负权、最短路）
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 BEFOREPOJ 3259 Wormholes(BellmanFord)
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! Each of