Wormholes

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.

题目大意：

John有n个农场，m条路径，w个虫洞。现在他想进行一个奇妙的旅行，使他能通过虫洞回到以前。问他能否实现。

一道Ballman—Ford模板题，判负环。

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define inf 0x3f3f3f
#define V 505
#define N 5205
struct node
{
    int u, v, t;
}e[N];
int dis[V];
int k,n;
int B_Ford(int s)
{
    for(int i=1;i<=n;i++)  //初始化
        dis[i]=inf;
    dis[s]=0;   //起点为0
    for(int i=1;i<n;i++)
    {
        int flag=1;
        for(int j=0;j<k;j++)
            if(dis[e[j].v]>dis[e[j].u]+e[j].t)
        {
            dis[e[j].v]=dis[e[j].u]+e[j].t;
            flag=0;
        }
        if(flag) return 0;  //如果不能松弛了，直接判断没有负环
    }
    for(int i=0;i<k;i++)
        if(dis[e[i].v]>dis[e[i].u]+e[i].t)
        return 1;   //如果还能松弛则存在负环
    return 0;
}
int main()
{
    int T, m, t, u, v, time;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d%d", &n, &m, &t);
        k = 0;
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d", &u, &v, &time);   //路径为双向
            e[k].u = u;
            e[k].v = v;
            e[k++].t = time;
            e[k].u = v;
            e[k].v = u;
            e[k++].t = time;
        }
        for(int i=0; i<t; i++)
        {
            scanf("%d%d%d", &u, &v, &time);  //虫洞为单向路径
            e[k].u = u;
            e[k].v = v;
            e[k++].t = -time;
        }
      if(B_Ford(1))
        printf("YES\n");
      else
        printf("NO\n");
    }
    return 0;
}