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.

题解：判断是否有负环。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define mem(a) memset(a,0,sizeof(a));

using namespace std;

const int INF = 0x3fffffff;

struct Node
{
	int from;
	int to;
	int time;
};

Node e[5505];
int d[550];

bool bellman(int n,int edge)
{
	for(int i = 1;i <= n;i++)
	{
		d[i] = INF;
	}
	d[1] = 0;         //找个起点 
	bool flag;
	for(int i = 1;i <= n;i++)
	{
		flag = false;
		for(int j = 0;j < edge;j++)
		{
			if(d[e[j].to] > d[e[j].from] + e[j].time)
			{
				flag = true;
				d[e[j].to] = d[e[j].from] + e[j].time;
			}
		}
		if(!flag)
		{
			return false;
		}
		if(i == n)
		{
			return true;
		}
	}
	
}

int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		int n,m,w;
		int k = 0;
		scanf("%d%d%d",&n,&m,&w);
		for(int i = 0;i < m;i++)
		{
			int u,v,c;
			scanf("%d%d%d",&u,&v,&c);
			e[k].from = u;
			e[k].to = v;
			e[k++].time = c;
			e[k].from = v;
			e[k].to = u;
			e[k++].time = c;
		}
		for(int i = 0;i < w;i++)
		{
			int u,v,c;
			scanf("%d%d%d",&u,&v,&c);
			e[k].from = u;
			e[k].to = v;
			e[k++].time = -c;
		}
		
		if(bellman(n,k))
		{
			printf("YES\n");
		}
		else
		{
			printf("NO\n");
		}
	}
	
	
	return 0;
 }