USACO 3.2 Magic Squares

USACO 3.2 Magic Squares

 BFS搜索题。代码写的比较乱，用一个set来存储已经访问过的结点。analysis中的标程有不少可以学习的地方，如几个操作的变换，用转换数组来写，代码清晰明了，encode方法就省去了set。我通过保存整个树来输出解，标程通过反向操作，来输出解，做法很巧妙。

#include  < iostream >
#include  < fstream >
#include  < set >
#include  < queue >

using   namespace  std;

ifstream fin( " msquare.in " );
ofstream fout( " msquare.out " );

#ifdef _DEBUG
#define  out cout
#define  in cin
#else
#define  out fout
#define  in fin
#endif

int  final[ 8 ];
set < int > visited;
char  result[ 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 + 1 ];

struct  queue_node{
     int  current[ 8 ];
    queue_node  * parent;
     char  op;
};

void  op( int   * current, char  c)
{
     int  tmp;
     switch (c){
         case   ' A ' :
             for ( int  i = 0 ;i < 4 ; ++ i)
                swap(current[i],current[ 7 - i]);
             break ;
         case   ' B ' :
            tmp  =  current[ 3 ];
             for ( int  i = 3 ;i >= 1 ; -- i)
                current[i]  =  current[i - 1 ];
            current[ 0 ]  =  tmp;
            tmp  =  current[ 4 ];
             for ( int  i = 4 ;i <= 6 ; ++ i)
                current[i]  =  current[i + 1 ];
            current[ 7 ]  =  tmp;
             break ;
         case   ' C ' :
            tmp  =  current[ 6 ];
            current[ 6 ]  =  current[ 5 ];
            current[ 5 ]  =  current[ 2 ];
            current[ 2 ]  =  current[ 1 ];
            current[ 1 ]  =  tmp;
             break ;
    }
}

int  cur_value( int   * cur)
{
     int  res  =   0 ;
     for ( int  i = 0 ;i < 8 ; ++ i){
        res *= 10 ;
        res += cur[i];
    }

     return  res;
}


void  solve()
{
     for ( int  i = 0 ;i < 8 ; ++ i){
         in >> final[i];
    }

    queue < queue_node *>  q;
    queue_node  * node  =   new  queue_node;
     for ( int  i = 0 ;i < 8 ; ++ i)
        node -> current[i]  =  i + 1 ;

    node -> parent  =  NULL;

    q.push(node);

     while (  ! q.empty() ){
        queue_node  * node  =  q.front();
        q.pop();

         int  cur  =  cur_value(node -> current);
         if ( visited.find( cur)  !=  visited.end())
             continue ;
        visited.insert(cur);

         bool  ok  =   true ;
         for ( int  i = 0 ;i < 8 ; ++ i){
             if (node -> current[i] != final[i]){
                ok  =   false ;
                 break ;
            }
        }

         if (ok){
             int  i  =   0 ;
             while (node -> parent != NULL){
                result[i ++ ]  =  node -> op;
                node = node -> parent;
            }

             if (i == 0 ){
                 out << 0 << endl << endl;
                exit( 0 );
            }

             out << i << endl;

             int  j;
            i -- ;
             for (j = 0 ;i >= 0 ;i -- ,j ++ ){
                 out << result[i];
                 if (j % 60 == 59 )  out << endl;
            }
            
             if (j % 60 != 0 )
                 out << endl;

            exit( 0 );
        }

         for ( char  c = ' A ' ;c <= ' C ' ; ++ c){
            queue_node  *  n  =   new  queue_node;
            memcpy(n -> current,node -> current, sizeof (node -> current));
            op(n -> current,c);
            n -> op  =  c;
            n -> parent  =  node;
            q.push(n);
        }
    }
}

int  main( int  argc, char   * argv[])
{
    solve(); 
     return   0 ;
}