rqnoj 70 八数码难题

rqnoj 70 八数码难题

Bfs + Hash is ok.
I found that the functions named mem... are veru useful.
Here is my code:

#include < iostream >
#include < string .h >
#define  maxn 1000000
using   namespace  std;
const   int  xd[] = { - 1 , 1 , 0 , 0 },yd[] = { 0 , 0 , - 1 , 1 };
int  st[maxn][ 3 ][ 3 ],front = 1 ,rear = 1 ,aim[ 3 ][ 3 ] = {{ 1 , 2 , 3 },{ 8 , 0 , 4 },{ 7 , 6 , 5 }},dist[maxn];
int  head[maxn],next[maxn];

long  hash( long  x)
{
     long  re = 0 ;
     for ( long  i = 0 ;i < 3 ;i ++ )
         for ( long  j = 0 ;j < 3 ;j ++ )
            re = (re * 10 + st[x][i][j]) % maxn;
     return  re;
}
void  insert( long  x)
{
     long  h = hash(x),p = head[h];
     while (p)
    {
         if (memcmp(st[p],st[x], sizeof (st[p])) == 0 )  return ;
        p = next[p];
    }
    next[x] = head[h];head[h] = x;
    rear ++ ;dist[rear] = dist[front] + 1 ;
}
long  bfs()
{
     while (front <= rear)
    {
         if (memcmp(st[front],aim, sizeof (aim)) == 0 )  return  front;
         //   Found answer , return the position of the final state
         long  x,y;
         bool  find_zero = false ;
         for (x = 0 ;x < 3 &&! find_zero;x ++ )
             for (y = 0 ;y < 3 &&! find_zero;y ++ )
                 if (st[front][x][y] == 0 )
                    find_zero = true ;
        x -- ;y -- ;
         //   Find the position of figure zero
         for ( long  k = 0 ;k < 4 ;k ++ )
        {
             long  newx = x + xd[k],newy = y + yd[k];
             if (newx >= 0 && newx < 3 && newy >= 0 && newy < 3 )
            {
                memcpy(st[rear + 1 ],st[front], sizeof (st[front]));
                st[rear + 1 ][x][y] = st[rear + 1 ][newx][newy];
                st[rear + 1 ][newx][newy] = 0 ;
                 //   Produce new states
                insert(rear + 1 );
                 //   Try to insert
            }
        }
        front ++ ;
    }
     return   - 1 ;
}
int  main()
{
     string  s;
    cin >> s;
     for ( long  i = 0 ;i < 3 ;i ++ )
         for ( long  j = 0 ;j < 3 ;j ++ )
            st[ 1 ][i][j] = s.at( 3 * i + j) - ' 0 ' ;
    dist[ 1 ] = 0 ;
     //   Input
     long  ans = bfs();
     //   Bfs
    cout << dist[ans] << endl;
     //   Output
return   0 ;
}