poj_3259_spfa

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 27177		Accepted: 9784

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.

<pre name="code" class="cpp">#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;

int n, m, w;
int map[510][510];
bool visit[510];
int path[510], in[510];

bool spfa()
{
	queue<int>q;
	visit[1] = 1;
	path[1] = 0;
	in[1]++;
	q.push(1);
	while (!q.empty())
	{
		int v = q.front();
		q.pop();
		if (in[v] >= n)
			return 1;
		visit[v] = 0;
		for (int i = 1; i <= n; i++)
		{
			if (path[v] + map[v][i] < path[i])
			{
				path[i] = path[v] + map[v][i];
				if (!visit[i])
				{
					visit[i] = 1;
				
					if (in[i] >= n)
						return 1;
					q.push(i);
					in[i]++;
				
				}
			} 
		}
		visit[v] = 1;
	}
	return 0;
}

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		memset(path, 0x3f, sizeof (path));
		memset(visit, 0, sizeof (visit));
		memset(in, 0, sizeof (in));
		memset(map, 0x3f, sizeof(map));
		scanf("%d%d%d", &n, &m, &w);
		for (int i = 0; i < m; i++)
		{
			int a, b, c;
			scanf("%d%d%d", &a, &b, &c);
			if (map[a][b] > c)
			{
				map[a][b] = c;
				map[b][a] = c;
			}
		}
		for (int i = 0; i < w; i++)
		{
			int a, b, c;
			scanf("%d%d%d", &a, &b, &c);
			c = -c;
			if (map[a][b] > c)
				map[a][b] = c;
		}
		if (spfa())
			printf("YES\n");
		else 
			printf("NO\n");
		
	}
	return 0;
}