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Wormholes
Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit 
Status 
Practice 
POJ 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

Hint

懒得自己打题意~

这道题是是求是否存在负权环。

题意 ： 一个famer有一些农场，这些农场里面有一些田地，田地里面有一些虫洞，田地和田地之间有路，虫洞有这样的性质： 时间倒流。问你这个农民能不能看到他自己，也就是说，有没有这样一条路径，能利用虫洞的时间倒流的性质，让这个人能在这个点出发前回去，这样他就是能看到他自己
解题思路：使用Bellman-Ford算法，看图中有没有负权环。有的话就是可以，没有的话就是不可以了。

输入N,M,W，表示N个点，M表示两点间无向，W表示两点间有向

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXN=10010;
const int INF=1<<30;
int dis[505];

struct edge
{
	int u,v,t;
}E[6000];

bool Bellman_Ford(int n, int m)
{
	int i,j;
	for(i=1; i<=n; i++) dis[i]=INF;
	dis[1]=0;
	for(i=1; i<=n-1; i++){
		for(j=0; j<m; j++){
			if(dis[E[j].u]+E[j].t < dis[E[j].v]) dis[E[j].v] = dis[E[j].u]+E[j].t;
			//cout<<E[j].u<<" "<<E[j].v<<" "<<dis[E[j].u]<<" "<<dis[E[j].v]<<endl;
		}
	}
	for(j=0; j<m; j++){
		if(dis[E[j].v] > dis[E[j].u]+E[j].t) return 1;
	}
	return 0;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int T;
	scanf("%d", &T);
	while(T--)
	{
		int n,m,w;
		scanf("%d%d%d", &n,&m,&w);
		int size=0;
		int i,j;
		for(i=1; i<=m; i++){
			int u,v,t;
			scanf("%d%d%d", &u,&v,&t);
			E[size].u=u;
			E[size].v=v;
			E[size].t=t;
			size++;
			E[size].v=u;
			E[size].u=v;
			E[size].t=t;
			size++;
		}
		for(i=1; i<=w; i++){
			int u,v,t;
			scanf("%d%d%d", &u,&v,&t);
			E[size].u=u;
			E[size].v=v;
			E[size].t=-t;
			size++;
		}
		int ok=Bellman_Ford(n,size);
		if(ok) printf("YES\n");
		else printf("NO\n");
		//cout<<m+w<<" "<<size<<endl;
	}
	return 0;
}