这道题是是求是否存在负权环。
题意 ： 一个famer有一些农场，这些农场里面有一些田地，田地里面有一些虫洞，田地和田地之间有路，虫洞有这样的性质： 时间倒流。问你这个农民能不能看到他自己，也就是说，有没有这样一条路径，能利用虫洞的时间倒流的性质，让这个人能在这个点出发前回去，这样他就是能看到他自己
解题思路：使用Bellman-Ford算法，看图中有没有负权环。有的话就是可以，没有的话就是不可以了。
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 20953 | Accepted: 7447 |
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 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
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
代码：
#include<iostream>
using namespace std;
struct node
{
int u,v,w;
}edge[6000];
int dis[505];
int n,m,w,index;
const int inf=0x7ffffff;
void add(int u,int v,int c)
{
index++;
edge[index].u=u;
edge[index].v=v;
edge[index].w=c;
}
bool bellman()
{
int u,v,w,i,j,flag;
for(i=1;i<=n;i++)
{
dis[i]=inf;
}
dis[1]=0;flag=0;
for(i=1;i<=n;i++)
{
for(j=1;j<=index;j++)
{
if(dis[edge[j].v]>dis[edge[j].u]+edge[j].w)
{
dis[edge[j].v]=dis[edge[j].u]+edge[j].w;
}
}
}
for(i=1;i<=index;i++)
if(dis[edge[i].v]>dis[edge[i].u]+edge[i].w) return true;
return false;
}
int main()
{
int i,j,t,u,v,c;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&w);
index=0;
for(i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&c);
add(u,v,c);
add(v,u,c);
}
for(i=0;i<w;i++)
{
scanf("%d%d%d",&u,&v,&c);
add(u,v,-1*c);
}
if(bellman()) printf("YES\n");
else printf("NO\n");
}
return 0;
}