POJ 3259 Wormholes(负权判断）

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本文探讨了在一个包含路径和特殊虫洞的农场中，是否存在一种循环路径使得旅行者能够回到过去的时间点。通过构建图论模型，并利用贝尔曼-福特算法检测是否存在负权回路，最终确定是否可以实现时间旅行。

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Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 56618		Accepted: 21131

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.

路是双向的，虫洞是单向的，只要存在负环，则一定满足条件。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
#include<string>
#include<vector>
const int INF=0x3f3f3f3f;
int N,M,W,d[502];
using namespace std;
struct node 
{
	int to,cost;
};
vector<node> rode[502];
void init()
{
	for(int i=0;i<=500;i++)
		d[i]=INF;
	d[1]=0;
}
bool Bell()
{
	bool bb;
	for(int i=0;i<N;i++)
	{
		bb=0;
		for(int j=1;j<=N;j++)
		{
			int L=rode[j].size();
			for(int k=0;k<L;k++)
			{
				node e=rode[j][k];
				if(d[e.to]>d[j]+e.cost)
				{
					d[e.to]=d[j]+e.cost;
					bb=1;
				}
			}
		}
		if(bb==0)return 0;
	}
	return 1;
}
int main()
{
	int T,i,j,k;scanf("%d",&T);
	while(T--)
	{
		init();
		scanf("%d%d%d",&N,&M,&W);
		for(i=1;i<=M+W;i++)
		{
			int x,y,z;scanf("%d%d%d",&x,&y,&z);
			if(i<=M)
			{
				node r;r.to=y,r.cost=z;
				rode[x].push_back(r);r.to=x;
				rode[y].push_back(r);//双向路
			}
			else
			{
				node r;r.to=y,r.cost=-z;
				rode[x].push_back(r);//单向虫洞
			}
		}
		if(Bell())printf("YES\n");
		else printf("NO\n");
		for(i=0;i<=N;i++)
			rode[i].clear();//一定记得清空，wa了好几次
	}
	return 0;
}