POJ 3259, Wormholes

Time Limit: 2000MS  Memory Limit: 65536K
Total Submissions: 6639  Accepted: 2341

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

---

//  POJ3259.cpp : Defines the entry point for the console application.
//

#include  < iostream >
using   namespace  std;

int  main( int  argc,  char *  argv[])
{
     int  cases;
    cin  >>  cases;

     int  paths[ 5200 ][ 3 ];
     int  d[ 501 ];

     int  N,M,W;
     int  S,E,T;
     for  ( int  c  =   0 ; c  <  cases;  ++ c)
    {
        fill( & d[ 0 ],  & d[ 501 ],  10000 );

        cin  >>  N  >>  M  >>  W;
         int  SIZE  =   0 ;

         for  ( int  i  =   0 ; i  <  M;  ++ i)
        {
            cin  >>  S  >>  E  >>  T;
            
            paths[SIZE][ 0 ]  =  S;
            paths[SIZE][ 1 ]  =  E;
            paths[SIZE][ 2 ]  =  T;
             ++ SIZE;
            paths[SIZE][ 0 ]  =  E;
            paths[SIZE][ 1 ]  =  S;
            paths[SIZE][ 2 ]  =  T;
             ++ SIZE;
        }

         for  ( int  i  =   0 ; i  <  W;  ++ i)
        {
            cin  >>  S  >>  E  >>  T;
            paths[SIZE][ 0 ]  =  S;
            paths[SIZE][ 1 ]  =  E;
            paths[SIZE][ 2 ]  =   - T;
             ++ SIZE;
        }

         for  ( int  i  =   0 ; i  <  N  -   1 ;  ++ i)
             for  ( int  j  =   0 ; j  <  SIZE;  ++ j)
                d[paths[j][ 1 ]]  =  min(d[paths[j][ 0 ]]  +  paths[j][ 2 ],d[paths[j][ 1 ]]);

         bool  avbl  =   false ;
         for  ( int  j  =   0 ; j  <  SIZE;  ++ j)
             if  (d[paths[j][ 1 ]]  >  d[paths[j][ 0 ]]  +  paths[j][ 2 ])
            {
                avbl  =   true ;
                 break ;
            };

         if  (avbl) cout  <<   " YES\n " ;
         else  cout  << " NO\n " ;
    }
     return   0 ;
}