Wormholes（负权回路）

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Problem 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 toF (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, <i>F</i>. <i>F</i> farm descriptions follow.< br>Line 1 of each farm: Three space-separated integers respectively: <i>N</i>, <i>M</i>, and <i>W</i>< br>Lines 2..<i>M</i>+1 of each farm: Three space-separated numbers (<i>S</i>, <i>E</i>, <i>T</i>) that describe, respectively: a bidirectional path between <i>S</i> and <i>E</i> that requires <i>T</i> seconds to traverse. Two fields might be connected by more than one path.< br>Lines <i>M</i>+2..<i>M</i>+<i>W</i>+1 of each farm: Three space-separated numbers (<i>S</i>, <i>E</i>, <i>T</i>) that describe, respectively: A one way path from <i>S</i> to <i>E</i> that also moves the traveler back <i>T</i> seconds.

 

Output

Lines 1..<i>F</i>: 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：给你n个点，m条路，最后的 num个黑洞路径，后面的m条路输入的是a-->b的距离，还有花费的时间，num个黑洞可以让你从a-->b倒回时间t,问你有没有可能在哪个点会遇到原来的自己。很有意思的一道题啊，典型的判断是不是负权回路啊。

这也是做判断是不是负权回路的第一个题，收钱二重循环一遍，然后在循环的时候发现还有dis[e]>dis[s]+x[e][s]的时候就存在赋权回路

#if 0
#include<iostream>
#include<cstring>
using namespace std;
int n,m,no,k;   //点，路，黑洞 
struct lu
{
	int a,b,t;
}mapp[10000];
int dis[10000];         

void ford_one()
{
		memset(dis,0x7f,sizeof(dis));
		dis[1]=0; 
		int a,b,t;
		for(int i=1; i<=n-1; i++)
		{	
			bool flag=0;
			for(int j=1; j<=k; j++)
			{
				a=mapp[j].a; 
				b=mapp[j].b;
				t=mapp[j].t;
				
				if(dis[b]>dis[a]+mapp[j].t)
				{
					dis[b]=dis[a]+mapp[j].t;
					flag=1;
				}
			}
			if(flag==0)
				break;
		}
}

int ford_two()
{
	int a,b,t;
	
	    
		for(int j=1; j<=k; j++)
		{
			a=mapp[j].a; b=mapp[j].b;t=mapp[j].t;
			if(dis[b]>dis[a]+mapp[j].t)
				return 1;
		}
	
	return 0;
}


int main()
{
	int T;

	cin>>T;
	while(T--)
	{
		
		cin>>n>>m>>no; 
		k=0;
		for(int i=1; i<=m; i++) 
		{
			int aa,bb,xx;
			cin>>aa>>bb>>xx;
			k++;
			mapp[k].a=aa;
			mapp[k].b=bb;
			mapp[k].t=xx;
			k+=1;
			mapp[k].a=bb;
			mapp[k].b=aa;
			mapp[k].t=xx;
		}
		
		for(int i=1; i<=no; i++)
		{
			int aa,bb,xx;
			cin>>aa>>bb>>xx;
			k+=1;
			mapp[k].a=aa;
			mapp[k].b=bb;
			mapp[k].t=-xx;
		 
		}
		
		ford_one();
		
		if(ford_two())
		{
			cout<<"YES"<<endl;
		}
		else
		{
			cout<<"NO"<<endl;
		}
	}
	
}
#endif