如何检查单元是否与现有单元垂直相邻(相邻)?

How to check if element is orthogonally adjacent (next to) to existing elements?(如何检查单元是否与现有单元垂直相邻(相邻)?)
本文介绍了如何检查单元是否与现有单元垂直相邻(相邻)?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在尝试制作一个简单的游戏,其中放置在嵌套列表中的建筑物必须紧挨着另一座建筑物。我面临的问题是,如果建筑物被放置在侧面,我不能使用for loop来检查现有建筑物,因为它不再是列表的一部分,因为建筑物的某些侧面什么都没有。

board = [['', '', '', ''],
         ['', '', '', ''],
         ['', '', '', ''],
         ['', '', '', '']]

building_list = ['HSE', 'FAC', 'SHP', 'HWY', 'BCH']

输出:

Turn 2
    A     B     C     D
 +-----+-----+-----+-----+
1|     |     |     |     |
 +-----+-----+-----+-----+
2|     | HSE |     |     |
 +-----+-----+-----+-----+
3|     |     |     |     |
 +-----+-----+-----+-----+
4|     |     |     |     |
 +-----+-----+-----+-----+
1. Build a HWY
2. Build a HWY
3. See remaining buildings
4. See current score

5. Save game
0. Exit to main menu
Your choice? 1
Build where? a3
You must build next to an existing building.

我当前的代码:

def build_buildings():
    global turn
    a = 0
    m = 1
    j = 1
    k = 1
    i = 1
    if option == 1:
        building = option1
    else:
        building = option2

    location = input('Build where? ')
    location.split()
    col = location[0]
    row = int(location[1]) - 1
    if col == 'a':
        col = 0
    elif col == 'b':
        col = 1
    elif col == 'c':
        col = 2
    else:
        col = 3
    if turn > 1:
        if col == 0:
            m = -1
        if row == 0:
            j = 0
        if col == 3:
            k = -1
        if row == 3:
            i = 0

        if (board[int(row + i)][col] or board[int(row - j)][col] or board[int(row)][col + k]
                or board[int(row)][col - m]) != '':
            board[int(row)][col] = building
        else:
            board[int(row)][col] = ''
            print('You must build next to an existing building.')
            print()

            turn -= 1
    else:
        board[int(row)][col] = building

推荐答案

尝试此代码:

def build(place: str, what: int, first_build: bool = False):
    if len(place) == 2 and 0 <= what < len(building_list):  # check the arguments
        x, y = x_axe.get(place[0].upper(), -1), int(place[1]) if place[1].isdigit() else -1  # get the x,y coordinates
        # check the conditions to build
        if 0 <= x <= width and 0 <= y <= height and board[y][x] == '' and 
                (first_build or any(
                    0 <= x + dx <= width and 0 <= y + dy <= height and board[y + dy][x + dx] != '' for dy, dx in
                    ((-1, 0), (0, 1), (1, 0), (0, -1)))):
            board[y][x] = building_list[what]  # build!


# global variables
board = [['', '', '', ''],
         ['', '', '', ''],
         ['', '', '', ''],
         ['', '', '', '']]
building_list = ['HSE', 'FAC', 'SHP', 'HWY', 'BCH']
x_axe = {'A': 0, 'B': 1, 'C': 2, 'D': 3}  # make dict with x indexes for use in build()
# get the max coordinates of the board
height = len(board) - 1
width = len(board[0]) - 1

# simple examples; replace with your code
build('a3', 0, True)  # first build with True as third argument
build('b3', 1)  # other buildings; will be built
build('d3', 2)  # will not be built because of conditions
build('B2', 2)  # will be built
build('S9', 3)  # will not be built due to wrong symbol and wrong y coordinate

# output the board. first, make the format strings for the table
line = f"  {'+-----' * (width + 1)}+"  # +-----+-----+-----+-----+
row = f" {'|{:^5}' * (width + 1)}|"  # |{:^5}|{:^5}|{:^5}|{:^5}|
header = f"   {'{:^6}' * (width + 1)}"  # {:^6}{:^6}{:^6}{:^6}
print(header.format(*x_axe.keys()), line, sep='
')  # header
for i in range(height + 1):
    print(str(i) + row.format(*board[i]), line, sep='
')

输出:

     A     B     C     D   
  +-----+-----+-----+-----+
0 |     |     |     |     |
  +-----+-----+-----+-----+
1 |     |     |     |     |
  +-----+-----+-----+-----+
2 |     | SHP |     |     |
  +-----+-----+-----+-----+
3 | HSE | FAC |     |     |
  +-----+-----+-----+-----+

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