有邻接表和堆优化的【最短路】

有邻接表和堆优化的【最短路】

http://acm.hdu.edu.cn/showproblem.php?pid=2544

#include < stdio.h >
#define  Max 1000001
#include < string >
struct  Road{
     int  dis;
     int  to;
     struct  Road  *  next;
};
Road map[ 1000001 ], * tail[ 1000001 ];
struct  Heap{
     int  x,dis;
}hh[ 1000001 ];
bool  hash[ 1000001 ];
int  n,m;
int  len;
void  PutHeap(Heap a)
{
     int  i;
     for (i =++ len;hh[i >> 1 ].dis  >  a.dis;i >>= 1 )
        hh[i]  =  hh[i >> 1 ];
    hh[i]  =  a;
}
Heap GetHeap()
{
    Heap min  =  hh[ 1 ];
    Heap last  =  hh[len -- ];
     int  i,s;
     for (i = 1 ; 2 * i <= len;i = s)
    {
        s  =  i << 1 ;
         if (s + 1 <= len  &&  hh[s + 1 ].dis  <  hh[s].dis)
            s  ++ ;
         if (hh[s].dis  <  last.dis)
            hh[i]  =  hh[s];
         else
             break ;
    }
    hh[i]  =  last;
     return  min;
}
int  bfs()
{
    Heap a;
    a.x  =   1 ;
    a.dis  =   0 ;
    PutHeap(a);
     while (len )
    {
        Heap min  =  GetHeap();
         if (min.x  ==  n)
             return  min.dis;
         int  x  =  min.x;
         if ( ! hash[x])
             continue ;
        hash[x]  =   false ;
         int  dis  =  min.dis;

        Road  * p;
        p  =  map[x].next;
         while (p != NULL)
        {
             int  b  =  p -> to;
             if (hash[b])
            {
                a.x  =  b;
                a.dis  =  p -> dis  +  dis;
                PutHeap(a);
            }
            p  =  p -> next;
        }
    }
}
int  main()
{
     while (scanf( " %d%d " , & n, & m),n + m)
    {
        len  =   0 ;
         for ( int  i = 1 ;i <= n;i ++ )
        {
            hash[i]  =   true ;
            map[i].next  =  NULL;
            tail[i]  =   & map[i];
        }
         for (i = 0 ;i < m;i ++ )
        {
             int  a,b,dis;
            Road  * p, * q;
            p  =  (Road  * )malloc( 12 );
            q  =  (Road  * )malloc( 12 );

            scanf( " %d%d%d " , & a, & b, & dis);
            p -> next  =  NULL;
            p -> dis  =  dis;
            p -> to  =  b;
            tail[a] -> next  =  p;
            tail[a]  =  tail[a] -> next;

            q -> next  =  NULL;
            q -> dis  =  dis;
            q -> to  =  a;
            tail[b] -> next  =  q;
            tail[b]  =  tail[b] -> next;;
        }
        printf( " %d\n " ,bfs());
    }
     return   0 ;
}