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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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：

农夫 FJ 有 N 块田地【编号 1...n】 (1<=N<=500)

        田地间有 M 条路径 【

双向】(1<= M <= 2500)

        同时有 W 个孔洞,可以回到以前的一个时间点【

单向】(1<= W <=200)

        问：FJ 是否能在田地中遇到以前的自己

算法：flod算法

思路： 田地间的双向路径加边,权值为正
        孔洞间的单向路径加边,权值为负【可以回到以前】
        判断有向图是否存在负环
      因为如果存在了负数环,时间就会不停的减少,
      那么 FJ 就可以回到以前更远的地方,肯定能遇到以前的自己的

#include
     
      #include
      
       #include
       
        using namespace std;const int maxn = 510;const int inf = 9999999;struct Node{    int u,v;    int t;}E[2500*2+200+10];int d[maxn];int c;int n,m,w;bool flod(){    for(int i=1;i
        
         d[u]+t){		//g跟新每个点时间                 d[v]=d[u]+t;                flag = true;            }        }        if(!flag) return false;		//如果不能跟新则不存在负边     }    for(int i=0;i
         
          d[E[i].u]+E[i].t) return true;    }    return false;}int main(){    int f;    scanf("%d",&f);    while(f--){        scanf("%d%d%d",&n,&m,&w);        c=0;        for(int i=0;i