TheLuckyString (SRM 428 Div2 500)

TheLuckyString　(SRM 428 Div2 500)

回溯生成所有的LuckyString，计数即可。10!*26操作数。在tc上不会超时。
看Summary中很多人用next_permutation生成所有全排列再判断是否为LuckyString，这样更快更方便一些。

    class  TheLuckyString
              { 
               public : 
                   int  len;
                   int  res;
                   int  cnt[ 26 ];
                   int  last;
                   int  visited[ 11 ];
               int  count( string  s) 
                  {                         
                        memset(cnt, 0 , sizeof (cnt));

                        len  =  s.size();
                        res  =   0 ;

                         for ( int  i = 0 ;i < s.size(); ++ i)
                            cnt[s[i] - ' a ' ] ++ ;

                         int  len  =  s.size();

                        bt();

                         return  res;

                         return   0 ;
                  } 

               void  bt(){
                   for ( int  i = 0 ;i < 26 ; ++ i){
                       if (cnt[i] != 0 ){
                          visited[ 0 ] = i;
                          cnt[i] -- ;
                          _bt( 1 );
                          cnt[i] ++ ;
                      }
                  }
              }

               void  _bt( int  depth){
                   if (depth == len){
                    res ++ ;
                     return ;
                  }                  
                       for ( int  i = 0 ;i < 26 ; ++ i){
                           if (cnt[i] != 0 ){
                               if (i == visited[depth - 1 ])
                                   continue ;
                            visited[depth]  =  i;
                            cnt[i] -- ;
                            _bt(depth + 1 );
                            cnt[i] ++ ;
                          }
                      }
              }
｝

next_permutation解：

    class  TheLuckyString
              { 
               public : 
               int  count( string  s) 
                  {                         

                   int  res  =   0 ;

                  sort(s.begin(),s.end());

                   do {
                       bool  ok  =   true ;
                       for ( int  i = 0 ;i + 1 < s.size(); ++ i){
                           if (s[i] == s[i + 1 ]){
                              ok  =   false ;
                               break ;
                          }
                      }
                      
                       if (ok)
                        res ++ ;

                  } while ( next_permutation(s.begin(),s.end()) );

                     return  res;
                  }
              }