POJ 3259 Wormholes

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本文探讨了一种特殊情况下实现时间旅行的可能性，通过分析路径和虫洞的特性，使用Bellman-Ford算法检测是否存在一条路径让旅行者回到过去。

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Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 32555		Accepted: 11831

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

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：约翰希望从他的某块地出发并且回到这块地，使得他回来的时间早于出发的时间，两块地之间有路径或虫洞，路径是双向的，虫洞是单向的，虫洞可以让时光倒流。

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

const int INF=0x3f3f3f3f;
int sum;
int n,m,ss;
int dis[510];

struct Edge
{
	int u;
	int v;
	int w;
}edge[5300];

int BF()
{
	for(int i=0;i<=n;i++)//手残到这种程度也真的是醉了
		dis[i]=INF;
	dis[0]=0;
	for(int i=1;i<n;i++)
	{
		int flag=0;
		for(int j=0;j<sum;j++)
		{
			if(dis[edge[j].v]>dis[edge[j].u]+edge[j].w)
			{
				dis[edge[j].v]=dis[edge[j].u]+edge[j].w;
				flag=1;
			}
		}
		if(!flag)//如果某次迭代中没有任何一个dis值改变，说明该遍迭代中的所有边均未被松弛，至此后，所有的边都不需再松弛，可直接退出迭代
			break;
	}
	for(int i=0;i<sum;i++)
	{
		if(dis[edge[i].v]>dis[edge[i].u]+edge[i].w)
			return 1;
	}
	return 0;
}

int main()
{
	int T;
	//int n,m,w;
	int s,e,t;
	scanf("%d",&T);
	while(T--)
	{
		sum=0;
		scanf("%d%d%d",&n,&m,&ss);
		for(int i=0;i<m;i++)
		{
			scanf("%d%d%d",&s,&e,&t);//路径是双向的
			edge[sum].u=s;
			edge[sum].v=e;
			edge[sum].w=t;
			sum++;
			edge[sum].u=e;
			edge[sum].v=s;
			edge[sum].w=t;
			sum++;
		}
		for(int i=0;i<ss;i++)
		{
			scanf("%d%d%d",&s,&e,&t);//虫洞是单向的
			edge[sum].u=s;
			edge[sum].v=e;
			edge[sum].w=-t;//虫洞的作用是时间倒流，因此要把权值设成负值
			sum++;
		}
		if(BF())
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}