题目：
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.
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.
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 8Sample Output
NO YES
题意：
大意，给出几个点，其中距离为正为负的已经分别给出，让我们判断是否有负环存在
代码：
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
const int inf=(1<<30);
int f,n,m,w,total;
int s[5500],e[5500],t[5500];
void read()
{
int i;
total=1;
int a,b,c;
scanf("%d %d %d",&n,&m,&w);
while(m--)
{
scanf("%d %d %d",&a,&b,&c);
s[total]=a;
e[total]=b;
t[total++]=c;
s[total]=b;
e[total]=a;
t[total++]=c;
}
while(w--)
{
scanf("%d %d %d",&a,&b,&c);
s[total]=a;
e[total]=b;
t[total++]=-c;
}
}
bool bellman_ford()
{
int i,j;
int dis[5500];
for(i=1;i<=n;i++)
dis[i]=inf;
dis[1]=0;
for(i=1;i<n;i++)
{
for(j=1;j<=total;j++)
{
int x=s[j];
int y=e[j];
if(dis[x]<inf&&dis[y]>dis[x]+t[j])
dis[y]=dis[x]+t[j];
}
}
for(i=1;i<=total;i++)
if(dis[e[i]]>dis[s[i]]+t[i])
return true;
return false;
}
int main()
{
scanf("%d",&f);
while(f--)
{
read();
if(bellman_ford())
printf("YES\n");
else
printf("NO\n");
}
}