Wormholes
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
题意描述：
有n个点，m条双向正权边，k条单向负权边，判断从某个点出发后能否再回来，如果有负权回路就能回来
程序代码：
#include<stdio.h>
#include<string.h>
#define inf 99999999
int F();
void add(int u,int v,int w);
struct data{
int u;
int v;
int w;
}a[10010];
int dis[510],n,m,k,q;
int main()
{
int i,t,u,v,w,count;
scanf("%d",&t);
while(t--)
{
q=1;
scanf("%d%d%d",&n,&m,&k);
while(m--)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
while(k--)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,-w);
}
count=F();
if(count==0)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
int F()
{
int i,j;
for(i=1;i<=n;i++)
dis[i]=inf;
dis[1]=0;
for(i=1;i<=n-1;i++)
for(j=1;j<=q;j++)
if(dis[a[j].v]>dis[a[j].u]+a[j].w)
dis[a[j].v]=dis[a[j].u]+a[j].w;
for(j=1;j<=q;j++)
if(dis[a[j].v]>dis[a[j].u]+a[j].w)
return 0;
return 1;
}
void add(int u,int v,int w)
{
a[q].u=u;
a[q].v=v;
a[q].w=w;
q++;
}