我该如何解决这个 tromino 平铺问题？

Question

我正在尝试为 nxn 板解决这个 tromino 平铺问题。 鉴于 nxn 板中缺少正方形坐标 (MS) x 和 y，我们必须用“L”形瓷砖填充板的其余部分。

我设法获得了 2 x 2 板的输出。 但是，我坚持使用 4 x 4、8 x 8 等电路板尺寸的输出。

以下是我编写的递归调用自身以打印板的函数：

/* This program tiles with right trominoes an nxn
   board with one square missing, assuming that n
   is a power of 2.                                */

#include <stdio.h>
#include <stdlib.h>

// #define LR 0
// #define LL 1
// #define UR 3
// #define UL 2
// #define MS -1

const int Max = 256;

int board[Max][Max];

void tromino /* function to do tiling */
     ( int x_board,      /* x coordinate of board */
       int y_board,      /* y coordinate of board */
       int x_missing,    /* x coordinate of missing square */
       int y_missing,    /* y coordinate of missing square */
       int board_size);   /* size of board */

void show_Tromino(int size, int x_miss, int y_miss);

int main()
{
     int board_size,
         x_missing,  /* x coordinate of missing square */
         y_missing;  /* y coordinate of missing square */
     do {
          printf( "\n-------------------------------------" );
          printf( "\nEnter size of board (0 to quit): " );
          scanf( "%d", &board_size );
          if ( board_size ) {
               printf( "\nEnter coordinates of missing square: " );
               scanf( "%d%d", &x_missing, &y_missing );
               printf( "\n\nTiling\n" );
               // if(board[x_board][0] == x_missing && board[0][y_board] == y_missing)
               //  printf("MS\t");
               //board[x_missing][y_missing] = -1;
               tromino( 0, 0, x_missing, y_missing, board_size);
               show_Tromino(board_size, x_missing, y_missing);
          }
     } while ( board_size );

     return EXIT_SUCCESS;
}

void tromino( int x_board,     /* x coordinate of board */
              int y_board,     /* y coordinate of board */
              int x_missing,   /* x coordinate of missing square */
              int y_missing,   /* y coordinate of missing square */
              int board_size) /* size of board */

{
     int half_size = board_size/2, /* size of subboard */
         x_center,  /* x coordinate of center of board */
         y_center,  /* y coordinate of center of board */
         x_upper_left,  /* x coordinate of missing square in upper
                                                     left subboard */
         y_upper_left,  /* y coordinate of missing square in upper
                                                     left subboard */
         x_upper_right, /* x coordinate of missing square in upper
                                                    right subboard */
         y_upper_right, /* y coordinate of missing square in upper
                                                    right subboard */
         x_lower_right, /* x coordinate of missing square in lower
                                                    right subboard */
         y_lower_right, /* y coordinate of missing square in lower
                                                    right subboard */
         x_lower_left,  /* x coordinate of missing square in lower
                                                     left subboard */
         y_lower_left;  /* y coordinate of missing square in lower
                                                     left subboard */

     if ( board_size == 2 ) /* 2x2 board */
     {
       for(int i=board_size-1; i>=0; i--)
       {
         for(int j=0;j<board_size;j++)
         {
           if(i==x_missing && j==y_missing)
            //printf("MS\t");
          {  board[i][j] = -1;}
           //printf( "%d %d ", x_board + 1, y_board + 1 );
           /* find and print orientation of tromino */
         if(x_missing == x_board)
          {
            if(y_missing == y_board)
              // printf("UR\t");
              {board[i][j] = 2;}
            else
              // printf("LR\t");
              {board[i][j] = 0;}
          }

          else
          {
            if(y_missing == y_board)
              // printf("UL\t");
              {board[i][j] = 3;}
            else
              // printf("LL\t");
              {board[i][j] = 1;}
          }
       }
       printf("\n");
     }
     return;
    }

   // if(board_size>2)
   // {   /* compute x and y coordinates of center of board */
     x_center = x_board + half_size;
     y_center = y_board + half_size;


         if((x_missing < x_center) && (y_missing < y_center))
         {
           //printf("UR\t");
           x_upper_left = x_center - 1;
           y_upper_left = y_center;
           x_upper_right = x_center;
           y_upper_right = y_center;
           x_lower_left = x_missing;
           y_lower_left = y_missing;
           x_lower_right = x_center;
           y_lower_right = y_center - 1;

         }
         if((x_missing >= x_center) && (y_missing < y_center))
         {
           //printf("UL\t");
           x_upper_left = x_center - 1;
           y_upper_left = y_center;
           x_upper_right = x_center;
           y_upper_right = y_center;
           x_lower_left = x_center - 1;
           y_lower_left = y_center - 1;
           x_lower_right = x_missing;
           y_lower_right = y_missing;
         }
         if((x_missing >= x_center) && (y_missing >= y_center))
         {
           //printf("LL\t");
           x_upper_left = x_center - 1;
           y_upper_left = y_center;
           x_upper_right = x_missing;
           y_upper_right = y_missing;
           x_lower_left = x_center - 1;
           y_lower_left = y_center - 1;
           x_lower_right = x_center;
           y_lower_right = y_center - 1;
         }
         if((x_missing < x_center) && (y_missing >= y_center))
         {
           //printf("LR\t");
           x_upper_left = x_missing;
           y_upper_left = y_missing;
           x_upper_right = x_center;
           y_upper_right = y_center;
           x_lower_left = x_center - 1;
           y_lower_left = y_center - 1;
           x_lower_right = x_center;
           y_lower_right = y_center - 1;
         }
     /* tile the four subboards */
     tromino( x_board, y_board + half_size,
          x_upper_left, y_upper_left, half_size);
     tromino( x_board + half_size, y_board + half_size,
          x_upper_right, y_upper_right, half_size);
     tromino( x_board + half_size, y_board,
          x_lower_right, y_lower_right, half_size);
     tromino( x_board, y_board,
                    x_lower_left, y_lower_left, half_size);

}

void show_Tromino(int size, int x_miss, int y_miss)
{
  int bsize[Max][Max];
  for(int i=size-1; i>=0;i--)
  {
    for(int j=0; j<size;j++)
    {
      if(!(bsize[i][j] == -1 || (j==x_miss && i==y_miss)))
      {
        if(bsize[i][j] == 1)
        {
          printf("LL\t");
        }

        else if(bsize[i][j] == 2)
        {
          printf("UL\t");
        }

        else if(bsize[i][j] == 3)
        {
          printf("UR\t");
        }

        else if(bsize[i][j] == 0)
        {
          printf("LR\t");
        }
      }
      else if (bsize[i][j] == -1 || (j==x_miss && i==y_miss))
      {
        printf("MS\t");
      }
      bsize[i][j] += 1;
    }

    printf("\n");
  }
}

output1 : 当电路板尺寸为 2 x 2 时（这是正确的）

Enter size of board (0 to quit): 2

Enter coordinates of missing square: 0 1


Tiling
------
MS  LR  
LR  LR

output2 ：当电路板尺寸为 4 x 4（这是不正确的）时。 不过，我得到了 Missing Square(MS) 的正确位置。

Enter size of board (0 to quit): 4

Enter coordinates of missing square: 0 1


Tiling
------
LR  LR  LR  LR  
LR  LR  LR  LR  
MS  LR  LR  LR  
LR  LR  LR  LR

output2 ：当电路板尺寸为 4 x 4（这是预期的）时

Enter size of board (0 to quit): 4

Enter coordinates of missing square: 0 1

Tiling
------

UL UL UR UR
UL UR UR UR
MS LR UR LR 
LR LR LR LR

笔记：

• UL- 左上角
• UR- 右上角
• LL- 左下
• LR-右下
• MS- 失踪广场

Answer 1

问题是由于基本情况。 在求解 2x2 时，您忘记考虑偏移。 每次只更新[0,0],[0,1],[1,0],[1,1]。 其余的值（除了丢失的图块）都是垃圾值。

我相信您的编译器默认将所有数组值设置为零。 所以垃圾值导致值LR。

编辑：添加了更正的代码。 只需将输出映射到字母（我已经打印了数字）

/* This program tiles with right trominoes an nxn
   board with one square missing, assuming that n
   is a power of 2.                                */

#include <stdio.h>
#include <stdlib.h>

// #define LR 4
// #define LL 1
// #define UR 3
// #define UL 2
// #define MS -1

# define Max 256
int board[Max][Max];

void tromino /* function to do tiling */
     ( int x_board,      /* x coordinate of board */
       int y_board,      /* y coordinate of board */
       int x_missing,    /* x coordinate of missing square */
       int y_missing,    /* y coordinate of missing square */
       int board_size);   /* size of board */

void show_Tromino(int size, int x_miss, int y_miss);

int main()
{
     int board_size,
         x_missing,  /* x coordinate of missing square */
         y_missing;  /* y coordinate of missing square */
     do {
          printf( "\n-------------------------------------" );
          printf( "\nEnter size of board (0 to quit): " );
          scanf( "%d", &board_size );
          if ( board_size ) {
               printf( "\nEnter coordinates of missing square: " );
               scanf( "%d%d", &x_missing, &y_missing );
               printf( "\n\nTiling\n" );
               // if(board[x_board][0] == x_missing && board[0][y_board] == y_missing)
               //  printf("MS\t");
               //board[x_missing][y_missing] = -1;
               tromino( 0, 0, x_missing, y_missing, board_size);
               show_Tromino(board_size, x_missing, y_missing);
          }
     } while ( board_size );

     return EXIT_SUCCESS;
}

void tromino( int x_board,     /* x coordinate of board */
              int y_board,     /* y coordinate of board */
              int x_missing,   /* x coordinate of missing square */
              int y_missing,   /* y coordinate of missing square */
              int board_size) /* size of board */

{
     int half_size = board_size/2, /* size of subboard */
         x_center,  /* x coordinate of center of board */
         y_center,  /* y coordinate of center of board */
         x_upper_left,  /* x coordinate of missing square in upper
                                                     left subboard */
         y_upper_left,  /* y coordinate of missing square in upper
                                                     left subboard */
         x_upper_right, /* x coordinate of missing square in upper
                                                    right subboard */
         y_upper_right, /* y coordinate of missing square in upper
                                                    right subboard */
         x_lower_right, /* x coordinate of missing square in lower
                                                    right subboard */
         y_lower_right, /* y coordinate of missing square in lower
                                                    right subboard */
         x_lower_left,  /* x coordinate of missing square in lower
                                                     left subboard */
         y_lower_left;  /* y coordinate of missing square in lower
                                                     left subboard */

     if ( board_size == 2 ) /* 2x2 board */
     {
       //Theres too much of nesting for only 4 cases. Got rid of it
       if(x_board==x_missing && y_board==y_missing)
         board[x_board+1][y_board] = board[x_board][y_board+1]=board[x_board+1][y_board+1]=2;

       if(x_board+1==x_missing && y_board==y_missing)
         board[x_board][y_board]= board[x_board][y_board+1]=board[x_board+1][y_board+1]=3;

       if(x_board==x_missing && y_board+1==y_missing)
         board[x_board][y_board]= board[x_board+1][y_board] = board[x_board+1][y_board+1]=1;

       if(x_board+1==x_missing && y_board+1==y_missing)
         board[x_board][y_board]= board[x_board+1][y_board] = board[x_board][y_board+1]=0;
       return;
     }

   // if(board_size>2)
   // {   /* compute x and y coordinates of center of board */
     x_center = x_board + half_size;
     y_center = y_board + half_size;


         if((x_missing < x_center) && (y_missing < y_center))
         {
           //printf("UR\t");
           x_upper_left = x_center - 1;
           y_upper_left = y_center;
           x_upper_right = x_center;
           y_upper_right = y_center;
           x_lower_left = x_missing;
           y_lower_left = y_missing;
           x_lower_right = x_center;
           y_lower_right = y_center - 1;
           board[x_upper_left][y_upper_left]=2;     //CHANGE MADE. You forgot to tile the board
           board[x_upper_right][y_upper_right]=2;
           board[x_lower_right][y_lower_right]=2;

         }
         if((x_missing >= x_center) && (y_missing < y_center))
         {
           //printf("UL\t");
           x_upper_left = x_center - 1;
           y_upper_left = y_center;
           x_upper_right = x_center;
           y_upper_right = y_center;
           x_lower_left = x_center - 1;
           y_lower_left = y_center - 1;
           x_lower_right = x_missing;
           y_lower_right = y_missing;
           board[x_upper_left][y_upper_left]=3;     //CHANGE MADE. You forgot to tile the board
           board[x_upper_right][y_upper_right]=3;
           board[x_lower_left][y_lower_left]=3;
         }
         if((x_missing >= x_center) && (y_missing >= y_center))
         {
           //printf("LL\t");
           x_upper_left = x_center - 1;
           y_upper_left = y_center;
           x_upper_right = x_missing;
           y_upper_right = y_missing;
           x_lower_left = x_center - 1;
           y_lower_left = y_center - 1;
           x_lower_right = x_center;
           y_lower_right = y_center - 1;
           board[x_upper_left][y_upper_left]=1;     //CHANGE MADE. You forgot to tile the board
           board[x_lower_left][y_lower_left]=1;
           board[x_lower_right][y_lower_right]=1;
         }
         if((x_missing < x_center) && (y_missing >= y_center))
         {
           //printf("LR\t");
           x_upper_left = x_missing;
           y_upper_left = y_missing;
           x_upper_right = x_center;
           y_upper_right = y_center;
           x_lower_left = x_center - 1;
           y_lower_left = y_center - 1;
           x_lower_right = x_center;
           y_lower_right = y_center - 1;
           board[x_lower_left][y_lower_left]=0;  //CHANGE MADE. You forgot to tile the board
           board[x_upper_right][y_upper_right]=0;
           board[x_lower_right][y_lower_right]=0;
         }
     /* tile the four subboards */
     tromino( x_board, y_board + half_size,
          x_upper_left, y_upper_left, half_size);
     tromino( x_board + half_size, y_board + half_size,
          x_upper_right, y_upper_right, half_size);
     tromino( x_board + half_size, y_board,
          x_lower_right, y_lower_right, half_size);
     tromino( x_board, y_board,
                    x_lower_left, y_lower_left, half_size);

}

void show_Tromino(int size, int x_miss, int y_miss)
{
  for(int i=size-1;i>=0;i--)
  {
     for(int j=0; j<size;j++)
     {
       if(j==x_miss && i==y_miss)
        printf("-1");
       else
        printf("%d ",board[j][i]);
     }
    printf("\n");
  }
  //int bsize[Max][Max]; //CHANGE: WTH is bsize? All changes were made to board. Edited and changed them all

}

Answer 2

有一个dynamic recursion解决方案，还有一个类似的dynamic iteration解决方案，有助于避免输入非常大时由于深度递归调用而导致的堆栈溢出错误。

---

动态 - 递归

分析：

* for a 2*2 cell, could simply fill it,
* for a 4*4 cell,
    * divide it into 4 2*2 smaller squares,
    * first fill the 2*2 square that already has 1 cell filled,
    * now the 2*2 square at the center of the original square has 3 empty cells, just fill it with a tile,
    * now all the 3 remain 2*2 squares divided in previous step, has 1 cell filled,
    * then for each of the 3 remain 2*2 squares, fill with a tile,
    * 
* for a 8*8 cell,
    * divide it into 4 4*4 smaller squares,
    * then fill the 4*4 square that already has 1 cell filled, in similar way as a 4*4 input,
    * now the 2*2 square at the center of the original square has 3 empty cells, just fill it with a tile,
    * now all the 3 remain 4*4 squares divided in previous step, has 1 cell filled,
    * then for each of the 3 remain 4*4 squares, fill it in similar way as a 4*4 input,
    * 
* for a n*n cell,
    repeat divide & conquer,
    steps:
    * if n = 2,
        this is base, simple fill a tile,
    * else
        * divide it into 4 (n/2 * n/2) squares,
        * work on the one already with has a tile, by a recursive call,
        * then put a tile at center of input cell, to fill the (2*2) square in center,
        * then for each of the 3 (n/2 * n/2) squares,
            * work on it, by a recursive call,
            * 
        * 
    * 
*

---

动态 - 迭代

类似于上面的Dynamic - recursion ，但它使用循环来避免深度递归调用。

因此它是首选。

分析：

* if it's base case (2*2 square),
    it's the same as `Dynamic - recursion`,
* if it's not base case,
    * divide it into 4 (n/2 * n/2) squares, mark the one already has a cell filled as A,
    * then fill (2*2) square at the center of input square first, leave the cell belong to A empty,
        now all the 4 (n/2 * n/2) squares, has a single cell filled,
    * 
    * then loop the 4 (n/2 * n/2) squares, for each:
        * work on it, by a recursive call,
        * 
    * 
*

---

形状的表示

• 细胞

需要它的x和y位置。

因此，结构可能如下所示，在c ：

typedef struct {
    int x;
    int y;
} Cell;

• L瓷砖

可能有 4 个方向，标记为： 1, 2, 3, 4 ；
并且需要L瓦片中间单元格的坐标来定位，

因此，结构可能如下所示，在c ：

typedef struct {
    char direction; // 1, 2, 3, 4;
    Cell mid; // position of middle cell of tile,
} Tile;

• 正方形

需要左/上单元格来决定它的位置。
和边长决定它的大小。

因此，结构可能如下所示，在c ：

typedef struct {
    Cell start; // top/left cell,
    int size; // length of side,
} Square;

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