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

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.

题意：给出n个点和m条双向边且权值为正，w条单向边，权值为负。要求给定一个图，判断图中是否有负环。
题意：判断是否有环显然需要使用Bellman-Ford算法。因为本题只需判断是否有负环的存在，而不需求最短路，所以可令初始dis均为0，如果第n次松弛成功，则有负环；否则没有。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define inf 0x3f3f3f3f
const int maxn=550;

double dis[maxn];

int tot;//边的总数
int way[maxn*maxn][2];//存边的起点和终点
int value[maxn*maxn];//存所花费的时间
int n,m,w;
bool Bellman()
{
    for(int i=1;i<=n;i++)
		dis[i]=0;
	for(int i=0;i<=n;i++)//循环n-1次
	{
		bool flag=false;//优化
		for(int j=0;j<tot;j++)
		{
			if(dis[way[j][1]]>dis[way[j][0]]+value[j])
			{
				flag=true;
				dis[way[j][1]]=dis[way[j][0]]+value[j];
			}
		}
		if(!flag)return false;//无环
	}
	return true;
}


int main()
{
	int t;
	scanf("%d",&t);
	int a,b,c;
	while(t--)
	{
		scanf("%d%d%d",&n,&m,&w);
		tot=0;
		for(int i=0;i<m;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			way[tot][0]=a;
			way[tot][1]=b;
			value[tot]=c;
			tot++;
			way[tot][0]=b;
			way[tot][1]=a;
			value[tot]=c;
			tot++;
		}
		for(int i=0;i<w;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			way[tot][0]=a;
			way[tot][1]=b;
			value[tot]=-c;
			tot++;
		}
		int i;

		if(Bellman())
			printf("YES\n");
		else
			printf("NO\n");
	}
    return 0;
}