棋盘覆盖python代码

以下是一个使用分治算法实现棋盘覆盖的Python代码：

def chessboard(board, tr, tc, dr, dc, size, count):
    if size == 1:
        return count

    s = size // 2
    t = count
    # 左上角
    if dr < tr + s and dc < tc + s:
        count = chessboard(board, tr, tc, dr, dc, s, count)
    else:
        board[tr + s - 1][tc + s - 1] = t
        count += 1
        count = chessboard(board, tr, tc, tr+s-1, tc+s-1, s, count)

    # 右上角
    if dr < tr + s and dc >= tc + s:
        count = chessboard(board, tr, tc+s, dr, dc, s, count)
    else:
        board[tr + s - 1][tc + s] = t
        count += 1
        count = chessboard(board, tr, tc+s, tr+s-1, tc+s, s, count)

    # 左下角
    if dr >= tr + s and dc < tc + s:
        count = chessboard(board, tr+s, tc, dr, dc, s, count)
    else:
        board[tr + s][tc + s - 1] = t
        count += 1
        count = chessboard(board, tr+s, tc, tr+s, tc+s-1, s, count)

    # 右下角
    if dr >= tr + s and dc >= tc + s:
        count = chessboard(board, tr+s, tc+s, dr, dc, s, count)
    else:
        board[tr + s][tc + s] = t
        count += 1
        count = chessboard(board, tr+s, tc+s, tr+s, tc+s, s, count)

    return count

def print_board(board):
    for i in range(len(board)):
        for j in range(len(board[i])):
            print(board[i][j], end='\t')
        print()

size = 8  # 棋盘大小
board = [[0] * size for i in range(size)]  # 初始化棋盘
special_row = 3  # 特殊方格行坐标
special_col = 4  # 特殊方格列坐标
board[special_row][special_col] = -1  # 在特殊方格上放置障碍物
chessboard(board, 0, 0, special_row, special_col, size, 1)  # 开始棋盘覆盖
print_board(board)  # 输出结果

该代码中使用了递归的思想，每次将棋盘划分成四个子棋盘，然后对每个子棋盘进行递归处理。在处理每个子棋盘的时候，如果子棋盘中包含特殊方格，则在特殊方格上放置一个障碍物。每次递归完成后，返回放置的方块数量。最后输出棋盘即可。