poj 3259

                                                       

                                                       Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 28499		Accepted: 10302

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

题意：John的农场里field块地，path条路连接两块地，hole个虫洞，虫洞是一条单向路，不但会把你传送到目的地，而且时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。

思路：bellman_ford。由于存在负权边，Dijkstra便不能用了。题目简化下，就是看图中有没有负权环，有的话John可以无限次走这个环，使得时间一定能得到一个负值。所以有的存在负环话就是可以，没有的话就是不可以了。

 

#include <iostream>
#include <cstdio>
using namespace std;
#define fMax 505
#define eMax 5205
#define wMax 99999

struct A
{
    int sta,end,time;
}edge[eMax];

int point_num,edge_num,dict[fMax];

bool bellman_ford()
{
   for(int i=2;i<=point_num;i++)
      dict[i]=wMax;
   for(int i=1;i<=point_num;i++)
   {
       bool fals=0;
       for(int j=1;j<=edge_num;j++)
       {
           int v=edge[j].sta;
           int u=edge[j].end;
           int w=edge[j].time;
           if(dict[v]>dict[u]+w)
           {
               dict[v]=dict[u]+w;
               fals=1;
           }
       }
       if(!fals)
       break;
   }
   for(int j=1;j<=edge_num;j++)
       {
           int v=edge[j].sta;
           int u=edge[j].end;
           int w=edge[j].time;
           if(dict[v]>dict[u]+w)
           {
               dict[v]=dict[u]+w;
               return 0;
           }
       }
       return 1;
}

int main()
{
    int farm;
    scanf("%d",&farm);
    while(farm--)
    {
        int field,path,hole;
        scanf("%d %d %d",&field,&path,&hole);
        int s,e,t,k=0;
        for(int i=1;i<=path;i++)
        {
            scanf("%d %d %d",&s,&e,&t);
            edge[++k].sta=s;
            edge[k].end=e;
            edge[k++].time=t;
            edge[k].sta=e;
            edge[k].end=s;
            edge[k].time=t;
        }
        for(int i=1;i<=hole;i++)
        {
            scanf("%d %d %d",&s,&e,&t);
            edge[++k].sta=s;
            edge[k].end=e;
            edge[k].time=-t;
        }
        point_num=field;
        edge_num=k;
        if(!bellman_ford())
        printf("YES ");
        else
        printf("NO ");

    }
    return 0;

}