POJ 3259 Wormholes

POJ 3259 Wormholes

Bellman_Ford 算法，

利用Bellman_Ford算法判断是否存在负回路。如果存在负回路就一定可以回到（不停的走这个回路，直到赚到的负值够与回去的值抵消的），如果不存在一定不可以回去。
用cin500+ms， 用scanf  100+ms

#include < iostream >
#include < stdio.h >
using   namespace  std;
const   int  INF = 0x0fffffff ;
struct  type
{
        int  s,t;
        int  len;
}edge[ 30000 ];

int  n,m,w,nedge = 0 ;
int  d[ 505 ];
bool  Bellman_Ford()
{
      bool  isPossible = false ;
      for ( int  j = 1 ; j <= n; j ++ )d[j] = INF;
     d[ 1 ] = 0 ;
      int  u,v;
      bool  f;
      for ( int  j = 1 ; j <= n - 1 ; j ++ )
     {
             f = 0 ;
              for ( int  k = 1 ; k <= nedge; k ++ )
             {
                     u = edge[k].s; v = edge[k].t;
                      if (d[v] > d[u] + edge[k].len)
                                 d[v] = d[u] + edge[k].len;
                     f = 1 ;            
             }
              if (f == 0 ) return   false ;
     }
  /*前面n-1次循环可以求出无负回路时的最短路径，再来一次如果可以更新就一定有负回路。/*
       for ( int  k = 1 ; k <= nedge; k ++ )
             {
                     u = edge[k].s; v = edge[k].t;
                      if (d[v] > d[u] + edge[k].len)
                           {  d[v] = d[u] + edge[k].len;  return   1 ;}
                                
             }
     
     //  system("pause");
      return   0 ;
     
}

int  main()
{
     int  t;
    cin >> t;
     while (t -- )
    {
               int  i,j,u,v,value;
              scanf( " %d %d %d " , & n, & m, & w); // cin>>n>>m>>w;
              nedge = 0 ;
               for ( int  i = 1 ; i <= m; i ++ )
              {
                      scanf( " %d %d %d " , & u, & v, & value); // cin>>u>>v>>value;
                       ++ nedge;
                      edge[nedge].s = u;
                      edge[nedge].t = v;
                      edge[nedge].len = value;
                       ++ nedge;
                      edge[nedge].s = v;
                      edge[nedge].t = u;
                      edge[nedge].len = value;
              }     
                  
               for ( int  i = 1 ; i <= w; i ++ )
              {
                      scanf( " %d %d %d " , & u, & v, & value); // cin>>u>>v>>value;
                       ++ nedge;
                      edge[nedge].s = u;
                      edge[nedge].t = v;
                      edge[nedge].len =- value;
                      
                      /*  ++nedge;
                      edge[nedge].s=v;
                      edge[nedge].t=u;
                      edge[nedge].len=-value;
                       */
              }
              
               if (Bellman_Ford())
                         cout << " YES " << endl;
               else        cout << " NO " << endl;
     
    }
    system( " pause " );
     return   0 ;
}