package POJ; import java.util.*; public class POJ_3259 { static int f,n,m,w; public int from,to; public double rate,commisions; static int edges;//边数 static class edge{ public int from,to,se; edge(){} edge(int from,int to,int se){ this.from=from;this.to=to;this.se=se;...
首先这个可以用Floyd去跑一遍,然后遍历每个点看看有没有能得到负权的点,想看Floyd做法的可以看看这篇博客:POJ 3259 Wormholes(Floyd判负环),这篇博客是用spfa去跑的,从一个点出发,我们只需要看在这个图中有没有负环存在(就是一个环的权值和为负的),如果有的话,我们就可以以这个负环中的任意一点为起点,最终...
package pro.proclass.chaopengfan; import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.Arrays; import java.util.StringTokenizer;publicclassVjudgeN {staticclassEdge {ints;inte;intv;publicEdge(ints,inte,intv) {this.s =s;this.e =e;this.v =v; } }publicstaticv...
算法:bellman_ford 判断是否有负环 思路: 田地间的双向路径加边,权值为正 孔洞间的单向路径加边,权值为负【可以回到以前】 判断有向图是否存在负环 因为如果存在了负数环,时间就会不停的减少, 那么FJ 就可以回到以前更远的地方,肯定能遇到以前的自己的 bellman_ford AI检测代码解析 #ifndef ONLINE_JUDGE #pragma...
【POJ - 3259 】Wormholes(Bellman_Ford或spfa算法,判断有向图中是否存在负环),题干: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
poj 3259 Wormholes 判断负权值回路 Wormholes Time Limit:2000 MSMemory Limit:65536 KB 64-bit integer IO format:%I64d , %I64uJava class name:Main [Submit] [Status] [Discuss] Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is ...
POJ 3259 Wormholes (判负环) Wormholes Time Limit : 4000/2000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 6 Accepted Submission(s) : 2 Problem Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A ...
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) . ...
需要注意的是不是连通图,因此在做spfa的时候要将所有的点先入队一次 【代码】 AI检测代码解析 #include<cstdio>#include<cstring>usingnamespacestd;constintN=200010,inf=~0U>>2,M=200000;intx,S,T,time[N],q[N],size,h,t,n,m,W,ed,dis[N],in[N],nxt[N],w[N],v[N],g[N],u,e,cost...
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>#include<algorithm>#include<cstring>#include<cmath>usingnamespacestd;constintSIZE=11111;constintMAXLEN=1<<30;structsss{ints,e,v; ...