Wormholes

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.

AC代码：

# include <stdio.h>

# define N 505
# define M 25000
# define INF 10000000

struct Edge
{
    int u;
    int v;
    int w;
};

Edge edge[M];
int n, m, w;
int num;

bool Bellman_ford()
{
    int dis[N];
    int i, j, k;
    for (i = 0; i <= n; i++)
    {
        dis[i] = 0;
    }
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < num; j++)
        {
            if (dis[edge[j].v] > dis[edge[j].u] + edge[j].w)
            {
                dis[edge[j].v] = dis[edge[j].u] + edge[j].w;
                if (i == n-1)
                {
                    return true;
                }
            }
        }
    }
    return false;
}

int main(void)
{
    int f;
    scanf("%d", &f);
    while (f--)
    {
        int i;
        int s, e, t;
        scanf("%d %d %d", &n, &m, &w);
        num = 0;
        for (i = 0; i < m; i++)
        {
            scanf("%d %d %d", &s, &e, &t);
            s--;
            e--;
            edge[num].u = s;
            edge[num].v = e;
            edge[num++].w = t;
            edge[num].u = e;
            edge[num].v = s;
            edge[num++].w = t;
        }
        for (i = 0; i < w; i++)  //反向虫洞
        {
            scanf("%d %d %d", &s, &e, &t);
            s--;
            e--;
            edge[num].u = s;
            edge[num].v = e;
            edge[num++].w = -t;
        }
        if (Bellman_ford())
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    return 0;
}