Wormholes POJ - 3259

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.

题意：
有n个点，m条边，代表现在可以走的通路，比如从a到b和从b到a需要花费c时间，现在在地上出现了w个虫洞，虫洞的
意义就是你从a到b花费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前
思路：
把虫洞看成是一条负权路，求一个图中是否存在负权回路。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
int main()
{
    int i, j, n, m, check, t, h, flag, k;
    int dis[6010], u[6010], v[6010], w[6010];
    scanf("%d", &t);
    while(t--)
    {
        flag = 0;
        scanf("%d%d%d", &n, &m, &h);
        for(i = 1; i <= 2*m; i += 2)
        {
            scanf("%d%d%d", &u[i], &v[i], &w[i]);
            u[i+1] = v[i];
            v[i+1] = u[i];
            w[i+1] = w[i];
        }
        for(i = 2*m+1; i <= 2*m+h; i++)
        {
            scanf("%d%d%d", &u[i], &v[i], &w[i]);
            w[i] = -w[i];
        }

        for(i = 1; i <= n; i++)
            dis[i] = inf;
        dis[1] = 0;
        for(j = 1; j <= n-1; j++)
        {
            check = 1;
            for(i = 1; i <= 2*m+h; i++)
            {
                if(dis[v[i]] > dis[u[i]] + w[i])
                {
                    dis[v[i]]  = dis[u[i]] + w[i];
                    check = 0;
                }
            }
            if(check)
                break;
        }
        for(i = 1; i <= 2*m+h; i++)
            if(dis[v[i]] > dis[u[i]] + w[i])
            {
                flag = 1;
                break;
            }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}