Wormholes

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16897		Accepted: 5959

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 <iostream>
#include <fstream>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <set>


using namespace std;

const int INF = 1<<30 - 1;

int Dist[501];



struct Edge
{
    int start;
    int end;
    int length;
};

vector<Edge> Edges;

int main()
{
    


    int F;

    cin >> F;

    while( F-- )
    {    
        int N,M,W;

        cin >> N >> M >> W;

        //clear
        for( int i = 0 ; i  < N ; ++i )
        {
            Dist[i] = INF;
        }
        Edges.clear();

        for( int i = 0 ; i < M ; ++i )
        {
            int S , E , T;

            cin >> S >> E >> T;

            Edge edge1;
            edge1.start = S;
            edge1.end = E;
            edge1.length = T;
            Edges.push_back(edge1);
            Edge edge2;
            edge2.start = E;
            edge2.end = S;
            edge2.length = T;
            Edges.push_back(edge2);
        }

        for( int i = 0 ; i < W ; ++i )
        {
            int S , E , T;

            cin >> S >> E >> T;

            Edge edge;
            edge.start = S;
            edge.end = E;
            edge.length = -T;
            Edges.push_back(edge);        
        }    

        //bellman
        Dist[0] = 0;
        for( int it = 0 ; it < N -1 ; ++it )
        {
            for( int i = 0; i < Edges.size() ; ++i )
            {
                if ( Dist[Edges[i].start] + Edges[i].length < Dist[Edges[i].end] )
                {
                    Dist[Edges[i].end] = Dist[Edges[i].start] + Edges[i].length;
                }
            }
        }

        for( int i = 0; i < Edges.size() ; ++i )
        {
            if ( Dist[Edges[i].start] + Edges[i].length < Dist[Edges[i].end] )
            {
                goto YES;
            }
        }

        printf("NO\n");
        continue;

YES:
        printf("YES\n");
        continue;
    }



    return 0;
}