POJ 3259: Wormholes(Bellman-Ford)

http://poj.org/problem?id=3259

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

题意分析：

John有一个农场，农场上有双向道路和虫洞，利用虫洞可以回到以前一段时间，现在John想知道有没有一种方案可以回到他走这条路之前的时间。

解题思路：

把双向道路的通过时间当成权值，虫洞的返回的时间作为负权值，判断图中是否存在负环。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#define N 550
using namespace std;
struct edge{
	int u;
	int v;
	int w;
}a[N*20];

int dis[N];
int main()
{
	int T, n, m, w, i, j, k, inf=0x9f9f9f;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d%d%d", &n, &m, &w);
		for(i=1; i<=m+w; i++)
		{
			scanf("%d%d%d", &a[i].u, &a[i].v, &a[i].w);
			if(i>m)
				a[i].w=-a[i].w;
			else
			{
				a[i+w+m].u=a[i].v;
				a[i+w+m].v=a[i].u;
				a[i+w+m].w=a[i].w;
			}
			
		}
		memset(dis, inf, sizeof(dis));
		dis[1]=0;
		for(k=1; k<n; k++)
		{
			int temp=0;
			for(i=1; i<=2*m+w; i++)
				if(dis[a[i].u]>dis[a[i].v]+a[i].w)
				{
					temp=1;
					dis[a[i].u]=dis[a[i].v]+a[i].w;
				}
			if(temp==0)
				break; 
		}
		int temp=0;
		for(i=1; i<=2*m+w; i++)
			if(dis[a[i].u]>dis[a[i].v]+a[i].w)
			{
				temp=1;
				break;
			}
		if(temp==0)
			printf("NO
");
		else printf("YES
");
	}
	return 0;
}