Python程序分治算法求棋盘覆盖问题

棋盘覆盖问题是指在一个2^n * 2^n的棋盘上，恰好有一个方格与其他方格不同，现在要用L型骨牌覆盖整个棋盘，求覆盖方案。分治算法是一种将问题分解成多个子问题来解决的算法，可以用来解决棋盘覆盖问题。具体步骤如下：

1.将棋盘分成四个大小相等的子棋盘，其中一个子棋盘包含特殊方格。
2.用L型骨牌覆盖其他三个子棋盘，递归地解决这三个子棋盘的棋盘覆盖问题。
3.递归地解决包含特殊方格的子棋盘的棋盘覆盖问题。

下面是Python程序实现分治算法求解棋盘覆盖问题的代码：

def chessboard_cover(board, tr, tc, dr, dc, size, special_row, special_col):
    global tile
    if size == 1:
        return
    t = tile
    tile += 1
    s = size // 2
    # 判断特殊方格所在的子棋盘
    if special_row < tr + s and special_col < tc + s:
        chessboard_cover(board, tr, tc, special_row, special_col, s, special_row, special_col)
    else:
        board[tr + s - 1][tc + s - 1] = t
        chessboard_cover(board, tr, tc, tr + s - 1, tc + s - 1, s, tr + s - 1, tc + s - 1)
    if special_row < tr + s and special_col >= tc + s:
        chessboard_cover(board, tr, tc + s, special_row, special_col, s, special_row, special_col)
    else:
        board[tr + s - 1][tc + s] = t
        chessboard_cover(board, tr, tc + s, tr + s - 1, tc + s, s, tr + s - 1, tc + s)
    if special_row >= tr + s and special_col < tc + s:
        chessboard_cover(board, tr + s, tc, special_row, special_col, s, special_row, special_col)
    else:
        board[tr + s][tc + s - 1] = t
        chessboard_cover(board, tr + s, tc, tr + s, tc + s - 1, s, tr + s, tc + s - 1)
    if special_row >= tr + s and special_col >= tc + s:
        chessboard_cover(board, tr + s, tc + s, special_row, special_col, s, special_row, special_col)
    else:
        board[tr + s][tc + s] = t
        chessboard_cover(board, tr + s, tc + s, tr + s, tc + s, s, tr + s, tc + s)

# 测试代码
size = 8
board = [[0 for j in range(size)] for i in range(size)]
tile = 1
special_row = 6
special_col = 6
chessboard_cover(board, 0, 0, special_row, special_col, size, special_row, special_col)
for i in range(size):
    for j in range(size):
        print(board[i][j], end='\t')
    print()