Poj3259 wormhole solution

subject

link

http://poj.org/problem?id=3259

Literal description

Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 92250 Accepted: 33922
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

Ideas

This problem depends on wormholes to reduce time infinitely , We can use SPFA To judge the negative ring , It can also be used. Floyd To judge the negative ring .

Code implementation

#include<cstdio>
#include<string.h>
#include<cstring>
using namespace std;

const int maxn=500+10;
int f,n,m,w;
int g[maxn][maxn];
inline bool Floyd(){
    
	for(int k=1;k<=n;k++){
    
		for(int i=1;i<=n;i++){
    
			for(int j=1;j<=n;j++){
    
				if(g[i][j]>g[i][k]+g[k][j])g[i][j]=g[i][k]+g[k][j];
			}
			if(g[i][i]<0)return true;
		}
	}
	return false;
}
int main(){
    
	scanf("%d",&f);
	while(f--){
    
		memset(g,0x3f,sizeof(g));
		scanf("%d%d%d",&n,&m,&w);
		for(int i=1;i<=m;i++){
    
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);
			if(g[u][v]>w||g[v][u]>w)g[u][v]=g[v][u]=w;
		}
		for(int i=1;i<=w;i++){
    
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);
			g[u][v]=-w;
		}
		if(!Floyd())printf("NO\n");
		else printf("YES\n"); 
	}
	return 0;
}