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
  
判断是否有负环
  
再次熟悉了一下bellman_ford
   
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const long INF = 10000000;
struct ty
{
    long s, t;
    double w;
};
long t, n, m, w, e;
ty edge[6010];
double dist[1010];
void insertedge(long x, long y, double z, long k)
{
    //cout << x << ' ' << y<< ' ' << z <<endl;
    edge[k].s = x;
    edge[k].t = y;
    edge[k].w = z;
}
void init()
{
    e = 0;
    scanf("%d%d%d", &n, &m, &w);
    for (long i = 1; i <= m; i++)
    {
        long s, t, l;
        scanf("%d%d%d", &s, &t, &l);
        insertedge(s, t, l , ++e);
        insertedge(t, s, l , ++e);
    }
    for (long i = 1; i <= w; i++)
    {
        long s, t, l;
        scanf("%d%d%d", &s, &t, &l);
        insertedge(s, t, -l , ++e);
    }
}
bool bellman_ford()
{
    int i, j;
    for(i = 2; i <= n; i ++)
        dist[i] = INF;
    for(i = 1; i < n; i ++)
    {
        bool boo = true;
        for(j = 1; j <= e; j ++)
        {
            int u = edge[j].s;
            int v = edge[j].t;
            int w = edge[j].w;
            if(dist[v] > dist[u] + w)
            {
                dist[v] = dist[u] + w;
                boo = false;
            }
        }
        if(boo)  break;
    }
    for(i = 1; i <= e; i ++)
    {
        int u = edge[i].s;
        int v = edge[i].t;
        int w = edge[i].w;
        if(dist[v] > dist[u] + w)   return false;
    }
    return true;
}
int main()
{
    freopen("poj3259.in", "r", stdin);
    scanf("%d",&t);
    while (t--)
    {
        init();
        bool b = false;
        if(!bellman_ford())   printf("YES\n");
        else  printf("NO\n");
    }
    return 0;
}