POJ 2965, The Pilots Brothers' refrigerator

POJ 2965, The Pilots Brothers' refrigerator
POJ 1753, Flip Game

这两道题类似，可转化图论的最短路径问题：
顶点（Vertex）数：65536
边（Edge）数：65536 * 16 / 2
边的权值（Weight）：1
对于给定两个点， 求最短路径：
由于边的权值均为1， 所以适宜采用BFS + Mark Table, 时间复杂度为O(N), N为顶点数。

Description

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

 

Input

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

 

Output

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

 

Sample Input
-+--
----
----
-+--

 

Sample Output
6
1 1
1 3
1 4
4 1
4 3
4 4

 

Source
Northeastern Europe 2004, Western Subregion

---

//  POJ1753.cpp : Defines the entry point for the console application.
//

#include  < algorithm >
#include  < iostream >
#include  < sstream >
using   namespace  std;

template < typename Target, typename Source >
Target lexical_cast(Source arg){
    std::stringstream interpreter;
    Target result;
    interpreter << arg;
    interpreter >> result;
     return  result;
}

void  minSwitchs( int  cposition)
{
     if  (cposition  ==   0 )
    {
        cout  <<   " 0\n " ;
         return ;
    }

     const  unsigned  short  pattern[ 16 ]  =  { 0xf888 , 0xf444 , 0xf222 , 0xf111 , 0x8f88 , 0x4f44 , 0x2f22 , 0x1f11 ,
                                         0x88f8 , 0x44f4 , 0x22f2 , 0x11f1 , 0x888f , 0x444f , 0x222f , 0x111f  };
     const   int  MAX  =   65536 ;

     char  cnt[MAX];
    unsigned  short  door[MAX];
    unsigned  short  queuen[MAX];
    memset(cnt, - 1 , sizeof (cnt));
    cnt[cposition]  =   0 ;

    queuen[ 0 ]  =  cposition;
     int  front  =   - 1 ;
     int     rear  =   0 ;

    unsigned  short  cpos  =   0 ;
    unsigned  short  npos  =   0 ;
     while  (front  !=  rear)
    {
         if ( ++ front  >   65535 )front  %=   65536 ;
        cpos  =  queuen[front];

         for  ( int  i  =   0 ; i  <   16 ;  ++ i) 
        {
            npos  =  cpos  ^  pattern[i];;
             if  (cnt[npos]  ==   - 1 )
            {
                 if ( ++ rear  >   65535 )rear  %=   65536 ;
                queuen[rear]  =  npos;
                cnt[npos]  =  cnt[cpos]  +   1 ;
                door[npos]  =  i;
            }
             if  (npos  ==   0 )
            {
                 int  ppos  =  npos;

                 string  ret  =   "" ;
                 int  steps  =  cnt[npos];
                 while  (ppos  !=  cposition)
                {
                    ret  +=   " \n " ;
                    ret  +=  ((door[ppos] & 3 )  +   1 )  +   ' 0 ' ;
                    ret  +=   "   " ;
                    ret  +=  ((door[ppos] >> 2 )  +   1 )  +   ' 0 ' ;
                    ppos  =  ppos  ^  pattern[door[ppos]];
                }
                std::reverse(ret.begin(), ret.end());
                ret  =  lexical_cast < string ,  int > (steps)  +   " \n "   +  ret;
                cout  <<  ret;
                 return ;
            }
        }
    }
    cout  <<   " -1 " ;
}


int  main( int  argc,  char *  argv[])
{
     int  cposition  =   0 ;
     char  s[ 5 ];
    s[ 4 ] = ' \0 ' ;
     for ( int  i = 1 ;i <= 4 ;i ++ )
    {
        scanf( " %s " ,s);
         for ( int  j = 0 ;j <= 3 ;j ++ )
        {
            cposition  <<=   1 ;
             if (s[j] == ' + ' ) ++ cposition;
        }
    }

    minSwitchs(cposition);
     return   0 ;
}