Wormholes

 

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34934		Accepted: 12752

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.

这题使用Bellman-Ford算法

#include <iostream>
using namespace std;
struct farm {
    int S;
    int E;
    int T;
} f[5500];
int main() {
    int num;
    int N, M, W;
    cin >> num;
    int F[500];
    for (int i = 0; i < num; i++) {
        cin >> N >> M >> W;
        for (int j = 0; j < N; j++) {
            F[j] = 20000;
        }
        F[0] = 0;
        for (int j = 0; j < M; j++) {
            int a, b, c;
            cin >> a >> b >> c;
            f[2*j].S = a;
            f[2*j].E = b;
            f[2*j].T = c;
            f[2*j+1].S = b;
            f[2*j+1].E = a;
            f[2*j+1].T = c;

        }
        for (int j =2* M; j < 2*M + W; j++) {
            int a, b, c;
            cin >> a >> b >> c;
            f[j].S = a;
            f[j].E = b;
            f[j].T = 0 - c;
        }
        for (int j = 0; j < N-1; j++) {
            for (int k = 0; k < 2*M + W; k++) {
                if (F[f[k].E] > F[f[k].S] + f[k].T) {
                    F[f[k].E] = F[f[k].S] + f[k].T;
                }
            }
        }
        int flag = 0;
        for (int k = 0; k < 2*M + W; k++) {
            if (F[f[k].E] >F[f[k].S] + f[k].T) {
                F[f[k].E] = F[f[k].S] + f[k].T;
                flag=1;
                break;
            }
        }
        if(flag){
            cout<<"YES"<<endl;
        }else{
            cout<<"NO"<<endl;
        }
    }
    return 0;
}