USACO 2.1 The Castle

USACO 2.1 The Castle

用Floodfill来给每个module着色，统计一下各种颜色的module数目。
在拆墙的时候，由于要求最南和最西的，从下向上，从左到右找，这样就可以满足条件。
拆墙的时候，如果墙两边的颜色不一样，则相加看是否最大。比最大值大则更新相应信息。

#include  < iostream >
#include  < fstream >

using   namespace  std;

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

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

int  n,m;
int  c[ 50 * 50 ];
int  data[ 50 ][ 50 ];
int  colors[ 50 ][ 50 ];
int  color_cnt;

const   int  W  =   1 ;
const   int  N  =   2 ;
const   int  E  =   4 ;
const   int  S  =   8 ;

void  color( int  i, int  j, int  color);

void  solve()
{
     in >> m >> n;
    
    memset(c, 0 , sizeof ( 0 ));
    memset(colors, - 1 , sizeof (colors));

    color_cnt  =   0 ;

     for ( int  i = 0 ;i < n; ++ i)
         for ( int  j = 0 ;j < m; ++ j){
             in >> data[i][j];
        }

     for ( int  i = 0 ;i < n; ++ i)
         for ( int  j = 0 ;j < m; ++ j){
             if (colors[i][j] ==- 1 ){
                color(i,j,color_cnt ++ );
            }
        }

     out << color_cnt << endl;

     int  largest  =  INT_MIN;

     for ( int  i = 0 ;i < color_cnt; ++ i){
        largest  =  max(largest,c[i]);
    }


     out << largest << endl;

    largest  =  INT_MIN;
     int  walli,wallj;
     char  dir;

      for ( int  j = 0 ;j < m; ++ j)
         for ( int  i = n - 1 ;i >= 0 ; -- i)
        {
              if (i - 1 >= 0 ){
                 if ( colors[i][j] != colors[i - 1 ][j]){
                     if ( largest < c[colors[i][j]] + c[colors[i - 1 ][j]]){
                        largest = c[colors[i][j]] + c[colors[i - 1 ][j]];
                        walli  =  i;
                        wallj  =  j;
                        dir  =   ' N ' ;
                    }
                }
            }

             if (j + 1 < m){
                 if ( colors[i][j] != colors[i][j + 1 ]){
                     if (largest < c[colors[i][j]] + c[colors[i][j + 1 ]]){
                        largest  =  c[colors[i][j]] + c[colors[i][j + 1 ]];
                        walli  =  i;
                        wallj  =  j;
                        dir  =   ' E ' ;
                    }
                }
            }

       }

     out << largest << endl;
     out << walli + 1 << '   ' << wallj + 1 << '   ' << dir << endl;

}

void  color( int  i, int  j, int  clr)
{
    if (colors[i][j] !=- 1 )
       return ;

    colors[i][j]  =  clr; 
    c[clr] ++ ;

     if ((data[i][j] & S) == 0 && (i + 1 < n)){
      color(i + 1 ,j,clr); 
    }

     if ((data[i][j] & E) == 0 && (j + 1 < m)){
        color(i,j + 1 ,clr);
    }

     if ((data[i][j] & W) == 0 && (j - 1 >= 0 )){
        color(i,j - 1 ,clr);
    }

     if ((data[i][j] & N) == 0 && (i - 1 >= 0 )){
        color(i - 1 ,j,clr);
    }
}

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

Compiling...
Compile: OK

Executing...
 Test 1: TEST OK [0.000 secs, 2868 KB]
 Test 2: TEST OK [0.022 secs, 2868 KB]
 Test 3: TEST OK [0.000 secs, 2868 KB]
 Test 4: TEST OK [0.000 secs, 2868 KB]
 Test 5: TEST OK [0.000 secs, 2868 KB]
 Test 6: TEST OK [0.000 secs, 2872 KB]
 Test 7: TEST OK [0.000 secs, 2872 KB]
 Test 8: TEST OK [0.011 secs, 2868 KB]

All tests OK.