poj3259Wormholes【最短路(判负环)】

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37423		Accepted: 13770

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.

Source

USACO 2006 December Gold

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define INF 11000
using namespace std;
int dis[555];
struct ant{
	int start, end, time;
}p[5555];
int n,m,w;
int bellman_ford()
{
	int i,j,k;
	int con = 1,f = 1;
	memset(dis,INF,sizeof(dis));
	while(f)
	{
		f = 0;
		if(con++ > n)	return 1;
		for(i = 1; i <= m; i++)
		{
			if(dis[p[i].start]+p[i].time < dis[p[i].end])
			{
				dis[p[i].end]=dis[p[i].start]+p[i].time;
				f = 1;
			}
			if(dis[p[i].end]+p[i].time < dis[p[i].start])
			{
				f = 1;
				dis[p[i].start]=dis[p[i].end]+p[i].time; 
			}
		}
		for( ; i <= m+w;i++)
		{
			if(dis[p[i].start]-p[i].time < dis[p[i].end])
			{
				dis[p[i].end]=dis[p[i].start]-p[i].time;
				f = 1; 
			}
		}	
	}
	return 0;
}

int main()
{
	int i,j,k;
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d%d",&n, &m, &w);
		for(i = 1; i <= m+w; i++)
		{
			scanf("%d%d%d",&p[i].start, &p[i].end, &p[i].time);
		}
		if(bellman_ford())	puts("YES");
		else	puts("NO");
	}
	return 0;
}