POJ 3259 Wormholes (最短路)

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34302		Accepted: 12520

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.

 

 

 

以每个点为起点跑一次，检测是否有负环。

 #include <iostream>
 #include <cstdio>
 using    namespace    std;

 const    int    INF = 0xfffffff;
 const    int    SIZE = ;
 int    D[];
 int    N,M,W,F;
 struct    Node
 {
     int    from,to,cost;
 }G[SIZE];

 bool    Bellman_Ford(int);
 bool    relax(int,int,int);
 int    main(void)
 {
     scanf("%d",&F);
     while(F --)
     {
         scanf("%d%d%d",&N,&M,&W);
         int    i = ;
         while(i <  * M)
         {
             scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);
             i ++;
             G[i] = G[i - ];
             swap(G[i].from,G[i].to);
             i ++;
         }
         while(i < W + M * )
         {
             scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);
             G[i].cost = -G[i].cost;
             i ++;
         }
         bool    flag = true;
         for(int i = ;i <= N;i ++)
             if(!Bellman_Ford(i))
             {
                 puts("YES");
                 flag = false;
                 break;
             }
         if(flag)
             puts("NO");
     }

     return    ;
 }

 bool    Bellman_Ford(int s)
 {
     fill(D,D + ,INF);
     D[s] = ;
     bool    update;

     for(int i = ;i < N - ;i ++)
     {
         update = false;
         for(int i = ;i < M *  + W;i ++)
             if(relax(G[i].from,G[i].to,G[i].cost))
                 update = true;
         if(!update)
             break;
     }
     for(int i = ;i < M *  + W;i ++)
         if(relax(G[i].from,G[i].to,G[i].cost))
             return    false;
     return    true;
 }

 bool    relax(int from,int to,int cost)
 {
     if(D[to] > D[from] + cost)
     {
         D[to] = D[from] + cost;
         return    true;
     }
     return    false;
 }

Bellman_Ford