Question

如何获得矩形的周长？

在下面的示例中它可以工作，但如果我将x1，y1，x2，y2更改为不同的值或使板更大，我从wall函数获得的坐标是错误的。墙功能有什么问题？

board_width = 8
board_height = 8
board = []

for line in range(0, board_height):
    lst = []
    for i in range(0, board_width):
        lst.append("x")
    board.append(lst)

#         X  Y  X2 Y2
# rect = [2, 4,  5, 7]
x1 = 2
y1 = 4
x2 = 5
y2 = 7


def create_rect():
    # go through the tiles in the rectangle and make them passable
    for x in range(x1, x2):
        for y in range(y1, y2):
            board[y][x] = " "

create_rect()

for i in board:
    for x in i:
        print x,
    print
print

# x1 = 1    y1 = 2      x2 = 4      y2 = 5
# answer
# x,y
# top wall --------
# 3,1 / 3,2 / 3,3 / 3,4 / 3,5 /
# lef wall |    |
# 1,3 / 1,4 / 1,5 / 1,6 / 1,7 /
# rgh wall |    |
# 5,3 / 5,4 / 5,5 / 5,6 / 5,7 /
# bot wall --------
# 1,7 / 2,7 / 3,7 / 4,7 / 5,7 /


def walls(x1, y1, x2, y2):
    flst = []
    lst = []
    # top wall
    for i in range(x1-1, x2+1):
        lst.append((x1+1, i))
    flst.append(lst)
    lst = []

    # left wall
    for i in range(y1-1, y2+1):
        lst.append((x1-1, i))
    flst.append(lst)
    lst = []

    # right wall
    for i in range(y1-1, y2+1):
        lst.append((x2, i))
    flst.append(lst)
    lst = []

    # bottom wall
    for i in range(x1-1, x2+1):
        lst.append((i, y2))
    flst.append(lst)

    return flst

nlst = walls(x1, y1, x2, y2)

for i in nlst:
    print i

Answer 1

添加一些代码以直观地覆盖在主板上的墙壁：

for points in nlst:
    for x, y in points:
        v = board[y][x]
        v = '+' if v == '.' else '.'
        board[y][x] = v

然后在最后打印电路板。即使使用电路板尺寸（8 x 8）和x1，y1等（2, 4, 5, 7）的当前值，您也会看到墙壁不正确。

然后注释掉walls()的不同部分以隔离麻烦。换句话说，一次只装配一面墙，只用一面墙打印电路板。你会发现问题在于顶壁。它看起来像这样：

x x x x x x x x
x x x . x x x x
x x x . x x x x
x x x . x x x x
x x   .   x x x
x x   .   x x x
x x       x x x
x x x x x x x x

您可以大致简化walls()：

def walls(x1, y1, x2, y2):
    xs = range(x1 - 1, x2 + 1)
    ys = range(y1 - 1, y2 + 1)
    return [
        [(x, y1 - 1) for x in xs],  # top
        [(x, y2) for x in xs],      # lft
        [(x1 - 1, y) for y in ys],  # rgt
        [(x2, y) for y in ys],      # bot
    ]

    # Even better: return a dict-of-lists so the data
    # structure would be self documenting.
    # {'top': [...], 'bot': [...], etc}

如何获得在船上创建的矩形周长

1 个答案: