Wormholes


Time Limit: 2000MS

 

Memory Limit: 65536K

Total Submissions: 22566

 

Accepted: 8056


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.


Source


​​USACO 2006 December Gold​​



思路：本题就是判断是否有出现负权回路。用SPFA，看是否有顶点入队数超过顶点数，有则说明有负权回路。


#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int nn=510;
const int mm=3100;
const int oo=1e9;
class node
{
  public:int v,c;
};
vector<node >e[nn];
int n,m,w,dis[nn],id[nn];
bool vis[nn];
queue<int>q;
bool spfa(int x)
{
  memset(id,0,sizeof(id));
  memset(vis,0,sizeof(vis));
  for(int i=0;i<n;i++)
    dis[i]=oo;
  dis[x]=0;vis[x]=1;++id[x];
  q.push(x);int z;
  while(!q.empty())
  {
    z=q.front();q.pop();vis[z]=0;
    for(int i=0;i<e[z].size();i++)
    {
      if(dis[e[z][i].v]>dis[z]+e[z][i].c)
      {
        dis[e[z][i].v]=dis[z]+e[z][i].c;
        if(!vis[e[z][i].v])
        {
          vis[e[z][i].v]=1;++id[e[z][i].v];
          q.push(e[z][i].v);
          if(id[e[z][i].v]>n)
            return 1;
        }
      }
    }
  }
  return 0;
}
int main()
{
  int cas;
  while(cin>>cas)
  {
    while(cas--)
    { memset(e,0,sizeof(e));
      cin>>n>>m>>w;
      int a,b,c;node z;
      for(int i=0;i<m;i++)
      {
        cin>>a>>b>>c;a--;b--;
        z.v=a;z.c=c;
        e[b].push_back(z);z.v=b;
        e[a].push_back(z);
      }
      for(int i=0;i<w;i++)
      {
        cin>>a>>b>>c;a--;b--;
        z.v=b;z.c=-c;
        e[a].push_back(z);
      }
      bool flag=0;
      for(int i=0;i<n;i++)
        if(spfa(i))
           {
             flag=1;break;
           }
        if(flag)cout<<"YES\n";
        else cout<<"NO\n";
    }
  }
}