poj Wormholes最短路问题（bellman_ford）

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 toF (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 fromS toE 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.

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAX 10000
#define INF 1<<29
typedef struct Edge
{
    int x,y,cost;
}Edge;
Edge edge[MAX];
int d[MAX],n,m,w;
int bellman_ford()
{
    int i,j;
    for(i=1;i<=n;i++)
        d[i]=INF;
        d[1]=0;
        for(i=1;i<=n-1;i++)
            for(j=1;j<=2*m+w;j++)
                if(d[edge[j].y]>d[edge[j].x]+edge[j].cost)
                        d[edge[j].y]=d[edge[j].x]+edge[j].cost;
        bool flag=1;
        for(i=1;i<=2*m+w;i++)
            if(d[edge[i].y]>d[edge[i].x]+edge[i].cost)
            {
                 flag=0;
                 break;
            }

        return flag;
}
int main()
{
    int f;
    cin>>f;
    while(f--)
    {
        int i=1,k;
        cin>>n>>m>>w;
       for(i=1;i<=2*m;i++)
        {
            scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].cost);
            i++;
            edge[i].x=edge[i-1].y;
            edge[i].y=edge[i-1].x;
            edge[i].cost=edge[i-1].cost;
        }
       for(i=2*m+1;i<=2*m+w;i++)
        {
             scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].cost);
             edge[i].cost=-edge[i].cost;
        }
        if(bellman_ford()==1)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}