Wormholes （Floyd 负权回路）

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

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.

题目大意：

有n个点，m条普通的正权双向的路，w条负权单向的虫洞，问是否能从某个点出发走一些路回来时回到出发的点。

即判断是否存在负权回路。

解题思路:这个题用的是floyd，不过注意正权的边是双向的，负权的边是单向的。

代码；

#include<stdio.h>
#include<string.h>
# define inf 99999999
int map[510][510];
int book[510],dis[510];

int main()
{ 
	int p,n,m,i,j,k,w,a,b,c,f;
	int s[210],e[210],t[210];
	scanf("%d",&p);
	while(p--)
	{
		scanf("%d%d%d",&n,&m,&w);
		for(i=1;i<=n;i++)
			for(j=1;j<=n;j++)
			{
				if(i==j)
					map[i][j]=0;
				else
					map[i][j]=inf;
			}
		for(i=1;i<=m;i++)//输入正权边 
		{
			scanf("%d%d%d",&a,&b,&c);
			if(map[a][b]>c)
			{
				map[a][b]=c;
				map[b][a]=c;
			}
		}
		for(i=1;i<=w;i++)//输入负权边 
		{
			scanf("%d%d%d",&a,&b,&c); 
			map[a][b]=-c;
		}
		f=0;
		for(k=1;k<=n;k++)
		{
			for(i=1;i<=n;i++)
			{
				for(j=1;j<=n;j++)
					if(map[i][j]>map[i][k]+map[k][j])
					{
						map[i][j]=map[i][k]+map[k][j];
					}
				if(map[i][i]<0)
				{
					f=1;
					break;
				} 	
			}
			if(f==1)    //若找到就跳出，否则会超时 
				break;
		}
		if(f==1)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}