USACO 1.4 The Clocks

USACO 1.4 The Clocks

4次同一操作等效于没有操作，操作的顺序不影响结果。
用迭代加深搜索解(貌似效率不高)，代码如下：

#include  < iostream >
#include  < fstream >
#include  < string >

using   namespace  std;

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

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

// op[i][j]为操作i对第j个时钟增加的小时数
int  ops[ 9 ][ 9 ];

string  op[ 9 ]  =  {
     " ABDE " , " ABC " , " BCEF " , " ADG " , " BDEFH " , " CFI " , " DEGH " , " GHI " , " EFHI "
};

// clocks当前时钟的状态
int  clocks[ 9 ];
// 当前的操作序列
int  cur_ops[ 9 ];

void  pre()
{
    memset(ops, 0 , sizeof (ops) );
    memset(cur_ops, 0 , sizeof (cur_ops));

     for ( int  i = 0 ;i < 9 ; ++ i){
         for ( int  j = 0 ;j < op[i].size(); ++ j){
            ops[i][op[i][j] - ' A ' ]  =   3 ;
        }
    }

}

// 执行clocks操作序列
void  do_ops()
{
     for ( int  i = 0 ;i < 9 ; ++ i){
         if (cur_ops[i] != 0 )
             for ( int  j = 0 ;j < 9 ; ++ j){
                clocks[j] += ops[i][j] * cur_ops[i];
                clocks[j] %= 12 ;
            }
    }
}

// i为当前到达的操作编号
void  dfs( int  i,  int  cur_depth, int  max_dep)
{
     // 到达迭代加深搜索的最大深度或i超过最大
     if (cur_depth > max_dep || i > 9 )
         return ;    

     if (cur_depth == max_dep){
        
         // 保存当前clocks状态
         int  saved[ 9 ];
        memcpy(saved,clocks, sizeof (clocks));

        do_ops();

         bool  valid  =   true ;
         for ( int  i = 0 ;i < 9 ; ++ i){
             if (clocks[i] != 0 ){
                valid  =   false ;
                 break ;
            }
        }

         if (valid){
             // 时钟全为0，到达可行解
             bool  first  =   true ;
             for ( int  i = 0 ;i < 9 ; ++ i){
                 if (cur_ops[i] != 0 ){
                     int  t  =  cur_ops[i];
                     while (t -- ){
                         if (first){
                             out << i + 1 ;
                            first  =   false ;
                        } else {
                             out << "   " << i + 1 ;
                        }
                    }
                }
            }

             out << endl;
            exit( 0 );
        } else {
             // 恢复时钟
            memcpy(clocks,saved, sizeof (saved));
             return ;
        }
    }

     for ( int  inc = 3 ;inc >= 0 ;inc -- ){
         // 因为要求最小的序列，优先多分配当前i,再分配后续更大的号
       if (cur_depth + inc <= max_dep && cur_ops[i] + inc < 4 ){
          cur_ops[i] += inc;
          dfs(i + 1 ,cur_depth + inc,max_dep);
          cur_ops[i] -= inc;
      } 
    }
}

void  solve()
{
    pre();

     for ( int  i = 0 ;i < 9 ; ++ i){
         in >> clocks[i];
        clocks[i] %= 12 ;
    }


     for ( int  d  =   0 ;d <= 3 * 9 ; ++ d){
        dfs( 0 , 0 ,d);
    }
}

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