POJ 3259 Wormholes （Bellman_ford算法）

题目链接：http://poj.org/problem?id=3259

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 45077		Accepted: 16625

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 FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

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 :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow. 
Line 1 of each farm: Three space-separated integers respectively: N, M, and W 
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path. 
Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

Output

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

Hint

For farm 1, FJ cannot travel back in time. 
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

 

题目大意：有一个农场有n块田地，m条路连接任意两条路（无向），w个虫洞（单向）回到t秒之前，现在需要你判断她从1到n，再从n回到1是否能遇到离开前的自己

解题思路：由于题目的特殊性，只需要将穿越虫洞的时间存为负值然后用Bellman_ford算法判断是否存在负权值回路即可 如果存在说明能看到否则反之。

Bellman_ford算法 http://www.cnblogs.com/Jason-Damon/archive/2012/04/21/2460850.html

AC代码：

 #include <stdio.h>
 #include <string.h>
 #define inf 9999999
 int dis[];
 int p[][];
 int n,m,w,ans;
 struct edge
 {
     int x,y,z;
 }g[<<];
 void add(int s,int e,int t)
 {
     g[ans].x = s;
     g[ans].y = e;
     g[ans].z = t;
     ans ++;
 }
 bool Bellman_ford()
 {
     int i,j;
     for (i = ; i <= n; i ++)
         dis[i] = inf;
     dis[] = ;

     for (i = ; i < n; i ++)
         for (j = ; j < ans; j ++)
             if (dis[g[j].y] > dis[g[j].x]+g[j].z)
                 dis[g[j].y] = dis[g[j].x]+g[j].z;

     for (j = ; j < ans; j ++) //遍历所有的边
         if (dis[g[j].y] > dis[g[j].x]+g[j].z)
             return false;
     return true;
 }
 int main ()
 {
     int s,e,t,i,j,f;
     scanf("%d",&f);
     while (f --)
     {
         ans = ;
         scanf("%d%d%d",&n,&m,&w);
         for (i = ; i < m; i ++)
         {
             scanf("%d%d%d",&s,&e,&t);
             add(s,e,t);
             add(e,s,t);
         }
         for (i = ; i < w; i ++)
         {
             scanf("%d%d%d",&s,&e,&t);
             add(s,e,-t);
         }
         bool f = Bellman_ford();
         if (f)
             printf("NO\n");
         else
             printf("YES\n");
     }
     return ;
 }