F

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.

给你三个数n,m,w，分别表示有n个农场，m条路，w个黑洞，接下来m行分别有三个数s，e，val表示s到e有一条权值为val 的边，接下来的w行每行有三个数s，e，val表示s到e有一条权值为0-val的边，因为黑洞可以回到过去，所以其权值为负。现在问你给你这些点，求每组数据是否存在一条回路能使时间倒退。寻找一条负权边，因为负权边在松弛时可以无限松弛，所以只要判断是否存在一条可以无限松弛的负权边就可以了。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#define INF 0x3f3f3f3fusing namespace std;struct node{   int u;   int v;   int step;}e[121212121];int n, m, w;bool Bellman_Ford(int ans){   int dis[1212121];   memset(dis, INF, sizeof(dis));   for(int i=1;i<=n;i++)   {      bool flag = false;      for(int j=1;j<ans;j++)      {         if(dis[e[j].v]>dis[e[j].u] + e[j].step)         {           flag = true;           dis[e[j].v] = dis[e[j].u] + e[j].step;         }      }      if(flag==false)      {         break;      }   }   for(int i=1;i<ans;i++)//寻找无限松弛的负权边   {      if(dis[e[i].v]>dis[e[i].u] + e[i].step)      return true;//找到   }   return false;}int main(){   int T;   scanf("%d", &T);   while(T--)   {      scanf("%d %d %d", &n, &m, &w);      int ans = 1;//记录边数      for(int i=0;i<m;i++)      {      int a, b, c;      scanf("%d %d %d", &a, &b, &c);//双向        e[ans].u = a;        e[ans].v = b;        e[ans++].step = c;        e[ans].u = b;        e[ans].v = a;        e[ans++].step = c;      }      for(int i=0;i<w;i++)//负权      {         int a, b, c;         scanf("%d %d %d", &a, &b, &c);         e[ans].u = a;         e[ans].v = b;         e[ans++].step = 0 - c;      }      if(Bellman_Ford(ans))      cout<<"YES"<<endl;      else      cout<<"NO"<<endl;   }   return 0;}