Ural 1033 Labyrinth

Ural 1033 Labyrinth

Accepted

0.015

217 KB

1033. Labyrinth

Time Limit: 1.0 second
Memory Limit: 16 MB

Administration of the labyrinth has decided to start a new season with new wallpapers. For this purpose they need a program to calculate the square of the walls inside the labyrinth. This job is just for you!

The labyrinth is represented by a matrix N× N (3 ≤ N ≤ 33, you see, ‘3’ is a magic digit!). Some matrix cells contain a dot character (‘.’) that denotes an empty square. Other cells contain a diesis character (‘#’) that denotes a square filled by monolith block of stone wall. All squares are of the same size 3×3 meters.

The walls are constructed around the labyrinth (except for the upper left and lower right corners, which are used as entrances) and on the cells with a diesis character. No other walls are constructed. There always will be a dot character at the upper left and lower right corner cells of the input matrix.

Your task is to calculate the square of visible part of the walls inside the labyrinth. In other words, the square of the walls' surface visible to a visitor of the labyrinth. Note that there's no holes to look or to move through between any two adjacent blocks of the wall. The blocks are considered to be adjacent if they touch each other in any corner. See picture for an example: visible walls inside the labyrinth are drawn with bold lines. The height of all the walls is 3 meters.

Input

The first line of the input contains the single number N. The next N lines contain N characters each. Each line describes one row of the labyrinth matrix. In each line only dot and diesis characters will be used and each line will be finished with a new line character. There will be no spaces in the input.

Output

Your program should print to the output a single integer — the exact value of the square of the wallpaper needed.

Sample

input	output
5 ..... ...## ..#.. ..### .....	198

搜索题：注意的要从两个入口处搜索，防止中间断开了
wa了两次 找不出错误，search(i+1，j)写成search(i+1,j+1)；

#include < iostream >
#include < cstring >
using   namespace  std;

int   const  maxSize = 35 ;

class  ural1033
{
public :
    ural1033(){ size = 0 ; memset(f, 0 , sizeof  f); }
     void  input();
     void  print();
     void  search( int  i,  int  j);
     int  size;
     int  getn(){ return  N;}
private :
     char  a[maxSize][maxSize];
     bool  f[maxSize][maxSize];
     int  N;
};

void  ural1033::input()
{
     cin >> N;
      int  i,j;
      for (i = 1 ; i <= N; i ++ )
         for (j = 1 ; j <= N; j ++ )
            cin >> a[i][j];
     for (i = 2 ; i <= N + 1 ; i ++ )a[ 0 ][i] = ' # ' ;
     for (i = 2 ; i <= N + 1 ; i ++ )a[i][ 0 ] = ' # ' ;
     for (i = 1 ; i <= N - 1 ; i ++ )a[N + 1 ][i] = ' # ' ;
     for (i = 1 ; i <= N - 1 ; i ++ )a[i][N + 1 ] = ' # ' ;

}

void  ural1033:: search( int  i,  int  j)
{
     if (i < 1 || i > N || j < 1 || j > N || a[i][j] == ' # ' || f[i][j] == 1 ) return  ;
    f[i][j] = 1 ;
     if (a[i - 1 ][j] == ' # ' )size ++ ;
        else  search(i - 1 ,j);
     if (a[i][j - 1 ] == ' # ' )size ++ ;
        else  search(i,j - 1 );
     if (a[i][j + 1 ] == ' # ' )size ++ ;
        else  search(i,j + 1 );
     if (a[i + 1 ][j] == ' # ' )size ++ ;
        else  search(i + 1 ,j);
}

void  ural1033::print()
{
      for ( int  i = 0 ; i <= N + 1 ; i ++ ,cout << endl)
      for ( int  j = 0 ; j <= N + 1 ; j ++ )
     cout << a[i][j] << '   ' ;
     cout << endl << endl;
     
      for ( int  i = 0 ; i <= N + 1 ; i ++ ,cout << endl)
      for ( int  j = 0 ; j <= N + 1 ; j ++ )
     cout << f[i][j] << '   ' ;
}

int  main()
{
    ural1033 ural;
    ural.input();
    
    ural.search( 1 , 1 );
    ural.search(ural.getn(),ural.getn());
    
    cout << ural.size * 9 << endl;
    
    system( " pause " );
     return   0 ;
}