POJ 3259：Wormholes bellman_ford判定负环

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37906		Accepted: 13954

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.

正常的path是双向的，有一定的消耗时间。虫洞是单向的，能够让时间倒流一定时间。问FJ能否找到一条路径，能够遇见之前的那个自己。

说白了就是找负环。

Bellman_ford模板题，用来对每一条边都进行松弛，然后看最后结果是否依然能够松弛。如果还能松弛，说明有负环；如果不能松弛了，就是没有负环。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

struct E{
	int s;
	int e;
	int l;
}edge[5205];

int N,M,W,edge_num;
int dis[505];

void addedge(int start,int end,int len)
{
	edge_num++;

	edge[edge_num].s=start;
	edge[edge_num].e=end;
	edge[edge_num].l=len;
}

bool bellman_ford()
{
	int i,j;
	for(i=1;i<=N-1;i++)
	{
		int flag=0;
		for(j=1;j<=edge_num;j++)
		{
			if(dis[edge[j].e]>dis[edge[j].s]+edge[j].l)
			{
				flag=1;
				dis[edge[j].e]=dis[edge[j].s]+edge[j].l;
			}
		}
		if(flag==0)
			break;
	}
	for(j=1;j<=edge_num;j++)
	{
		if(dis[edge[j].e]>dis[edge[j].s]+edge[j].l)
		{
			return true;
		}
	}
	return false;
}

int main()
{
	int i,start,end,len;
	int Test;
	cin>>Test;

	while(Test--)
	{
		edge_num=0;
		memset(dis,0,sizeof(dis));

		cin>>N>>M>>W;

		for(i=1;i<=M;i++)
		{
			cin>>start>>end>>len;

			addedge(start,end,len);
			addedge(end,start,len);
		}
		for(i=1;i<=W;i++)
		{
			cin>>start>>end>>len;

			addedge(start,end,-len);
		}
		if(bellman_ford())
		{
			cout<<"YES"<<endl;
		}
		else
		{
			cout<<"NO"<<endl;
		}
	}
	return 0;
}

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