扫雷，显示空单元格和所有其他空单元格逻辑和 python
minesweeper, reveal empty cell and all other empty cell logic and python
我目前正在创建扫雷器，并且已经到了我必须创建一个函数的部分，一旦玩家点击一个空的单元格，它就会显示该单元格和所有其他空单元格，包括编号单元格包围空单元格的集群（来吧伙计们，你知道扫雷 :) 我编写这个游戏的方法是有一个包含值的列表列表，0 是空的，1-8 是接触那个单元格的炸弹数量， 50是炸弹。我称之为网格。然后我使用这个网格来绘制整个游戏。然后使用另一个列表列表隐藏该网格，这些列表都包含布尔值。真隐藏，假显明。我称之为 boolgrid。例如，当玩家点击 cell[2][10] 时，boolgrid[2][10] 变为 False，显示该 cell.
选择空单元格时，必须显示周围的单元格，在某些情况下，周围的单元格也是空的，因此需要显示该单元格的周围单元格，依此类推。我的问题是编写一个支持它的函数，我尝试了很多东西，比如创建一个单元格坐标元组列表，其中单元格 == 0，然后创建一个新列表来保存所有最终可用于的新元组将所有相应的 boolgrid 单元格设置为 False。它不能很好地工作，很乱，不符合 Python 标准，占用大量内存。
我将非常感谢任何能帮助我实现此功能的人。
下面是一些精简代码，其中包含基本元素。该代码在网格中包含全 0，因此每个布尔值都应变为 False
# used to hold values to denote what the cell contains,
grid = [[0 for x in range(30)] for x in range(15)]
# used to check what cell is hidden, true is hidden, false is revealed,
booleangrid = [[True for x in range(30)] for x in range(15)]
list_to_hold_tuples = []
def find_connected_0_value_cells(cell):
i = 5 # used as an example in the real program these are whatever cell the player selects
j = 10 # as above
# this function should be given an argument of a cell that the player selected.
# reveal the surrounding cells, AND THEN take each surrounding cell, and go through
# the same process again, until every surrounding cell containing 0, and its surrounding
# cells containing zero have been revealed.
# i know i need some kind of loop, but cannot seem to build it.
# currently this function just reveals the cell and its immediate surrounding cells
if cell[i][j] == 0:
s = i, j + 1
t = i, j - 1
u = i - 1, j - 1
v = i - 1, j
w = i - 1, j + 1
x = i + 1, j - 1
y = i + 1, j
z = i + 1, j + 1
tempholding = [s, t, u, v, w, x, y, z]
for i in tempholding:
list_to_hold_tuples.append(i)
def reveal_empty_cells():
for a,b in list_to_hold_tuples:
booleangrid[a][b] = False
print(booleangrid)
find_connected_0_value_cells(grid)
reveal_empty_cells()
好的，所以我对您的代码进行了一些工作，并且能够编写出有效的代码。我确实调整了网格大小只是为了更容易理解，但它应该适用于您的原始大小。只要确保您没有像发布的代码中那样使用全 0 的网格进行测试，因为它会 return 太多递归的错误。这是代码。
编辑：我更改了代码，所以现在它也显示了周围的数字
# used to hold values to denote what the cell contains,
grid = [[0,1,2,5,4,6,0,0,0,0],
[0,1,2,5,4,6,0,0,0,0],
[0,1,2,5,4,6,0,0,5,0],
[0,1,2,5,4,6,0,0,0,0],
[0,1,2,5,4,6,0,0,0,0],
[0,1,2,5,4,6,6,0,0,0]]
# used to check what cell is hidden, true is hidden, false is revealed,
booleangrid = [[True for x in range(10)] for x in range(6)]
list_to_hold_tuples = []
def find_connected_0_value_cells(i,j):
list_to_hold_tuples.append((i,j))
if grid[i][j] == 0:
s = (i, j + 1)
t = (i, j - 1)
u = (i - 1, j - 1)
v = (i - 1, j)
w = (i - 1, j + 1)
x = (i + 1, j - 1)
y = (i + 1, j)
z = (i + 1, j + 1)
tempholding = [s, t, u, v, w, x, y, z]
for a in tempholding:
if a[0]>=0 and a[1]>=0 and a[0]<=len(grid)-1 and a[1]<=len(grid[i])-1 and grid[a[0]][a[1]]==0:
if (a[0]!=i and a[1]!=j) and (a not in list_to_hold_tuples):
find_connected_0_value_cells(a[0],a[1])
list_to_hold_tuples.append(a)
elif a[0]>=0 and a[1]>=0 and a[0]<=len(grid)-1 and a[1]<=len(grid[i])-1:
list_to_hold_tuples.append(a) #this part adds surrounding non-zero numbers
def reveal_empty_cells():
global booleangrid
for a,b in list_to_hold_tuples:
if a>=0 and a<len(booleangrid) and b>=0 and b<len(booleangrid[a]):
booleangrid[a][b] = False
for i in booleangrid:
print(i)
find_connected_0_value_cells(5,8)#Just using coordinates of 5,8 for testing, this would be click spot
reveal_empty_cells()
我已经重构并让其余部分正常工作。
谢谢@zettatekdev
grid = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 1, 2, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 2, 2, 0, 0, 0, 0, 0, 0],
[1, 1, 2, 3, 0, 0, 0, 0, 0, 0],
[1, 1, 2, 5, 0, 0, 0, 0, 0, 0],
[1, 1, 2, 5, 0, 0, 0, 0, 0, 0]]
list_to_hold_tuples = []
list_change_boolgrid =[]
row = 6
cell = 10
booleangrid = [[True for x in range(cell)] for x in range(row)]
def find_connected_0_value_cells(a, b):
list_to_hold_tuples.append((a, b))
if grid[a][b] == 0:
coord_list = get_surrounding_coords(a, b)
for a,b in coord_list:
if check_coord_values(a, b):
if grid[a][b] != 0:
c = a,b
if c not in list_to_hold_tuples:
list_to_hold_tuples.append(c)
else:
c = a,b
if c not in list_to_hold_tuples:
find_connected_0_value_cells(a,b)
def add_surrounding_cells():
extra_coord = True
for a,b in list_to_hold_tuples:
if grid[a][b] == 0:
coord_list = get_surrounding_coords(a,b, extra_coord)
for i in coord_list:
if i not in list_change_boolgrid:
list_change_boolgrid.append(i)
else:
c = a,b
if c not in list_change_boolgrid:
list_change_boolgrid.append(c)
def reveal_empty_cells():
global booleangrid
for a, b in list_change_boolgrid:
if check_coord_values(a,b):
booleangrid[a][b] = False
def check_coord_values(a,b):
if a == -1 or a >= row:
return False
if b == -1 or b >= cell:
return False
else:
return True
def get_surrounding_coords(a, b, *extra_coord):
c = (a, b + 1)
d = (a, b - 1)
e = (a - 1, b - 1)
f = (a - 1, b)
g = (a - 1, b + 1)
h = (a + 1, b - 1)
i = (a + 1, b)
j = (a + 1, b + 1)
if extra_coord:
k = (a, b)
return [c, d, e, f, g, h, i, j, k]
return [c, d, e, f, g, h, i, j]
find_connected_0_value_cells(3,5)
add_surrounding_cells()
reveal_empty_cells()
print(booleangrid)
我目前正在创建扫雷器，并且已经到了我必须创建一个函数的部分，一旦玩家点击一个空的单元格，它就会显示该单元格和所有其他空单元格，包括编号单元格包围空单元格的集群（来吧伙计们，你知道扫雷 :) 我编写这个游戏的方法是有一个包含值的列表列表，0 是空的，1-8 是接触那个单元格的炸弹数量， 50是炸弹。我称之为网格。然后我使用这个网格来绘制整个游戏。然后使用另一个列表列表隐藏该网格，这些列表都包含布尔值。真隐藏，假显明。我称之为 boolgrid。例如，当玩家点击 cell[2][10] 时，boolgrid[2][10] 变为 False，显示该 cell.
选择空单元格时，必须显示周围的单元格，在某些情况下，周围的单元格也是空的，因此需要显示该单元格的周围单元格，依此类推。我的问题是编写一个支持它的函数，我尝试了很多东西，比如创建一个单元格坐标元组列表，其中单元格 == 0，然后创建一个新列表来保存所有最终可用于的新元组将所有相应的 boolgrid 单元格设置为 False。它不能很好地工作，很乱，不符合 Python 标准，占用大量内存。
我将非常感谢任何能帮助我实现此功能的人。
下面是一些精简代码，其中包含基本元素。该代码在网格中包含全 0，因此每个布尔值都应变为 False
# used to hold values to denote what the cell contains,
grid = [[0 for x in range(30)] for x in range(15)]
# used to check what cell is hidden, true is hidden, false is revealed,
booleangrid = [[True for x in range(30)] for x in range(15)]
list_to_hold_tuples = []
def find_connected_0_value_cells(cell):
i = 5 # used as an example in the real program these are whatever cell the player selects
j = 10 # as above
# this function should be given an argument of a cell that the player selected.
# reveal the surrounding cells, AND THEN take each surrounding cell, and go through
# the same process again, until every surrounding cell containing 0, and its surrounding
# cells containing zero have been revealed.
# i know i need some kind of loop, but cannot seem to build it.
# currently this function just reveals the cell and its immediate surrounding cells
if cell[i][j] == 0:
s = i, j + 1
t = i, j - 1
u = i - 1, j - 1
v = i - 1, j
w = i - 1, j + 1
x = i + 1, j - 1
y = i + 1, j
z = i + 1, j + 1
tempholding = [s, t, u, v, w, x, y, z]
for i in tempholding:
list_to_hold_tuples.append(i)
def reveal_empty_cells():
for a,b in list_to_hold_tuples:
booleangrid[a][b] = False
print(booleangrid)
find_connected_0_value_cells(grid)
reveal_empty_cells()
好的，所以我对您的代码进行了一些工作，并且能够编写出有效的代码。我确实调整了网格大小只是为了更容易理解，但它应该适用于您的原始大小。只要确保您没有像发布的代码中那样使用全 0 的网格进行测试，因为它会 return 太多递归的错误。这是代码。 编辑：我更改了代码，所以现在它也显示了周围的数字
# used to hold values to denote what the cell contains,
grid = [[0,1,2,5,4,6,0,0,0,0],
[0,1,2,5,4,6,0,0,0,0],
[0,1,2,5,4,6,0,0,5,0],
[0,1,2,5,4,6,0,0,0,0],
[0,1,2,5,4,6,0,0,0,0],
[0,1,2,5,4,6,6,0,0,0]]
# used to check what cell is hidden, true is hidden, false is revealed,
booleangrid = [[True for x in range(10)] for x in range(6)]
list_to_hold_tuples = []
def find_connected_0_value_cells(i,j):
list_to_hold_tuples.append((i,j))
if grid[i][j] == 0:
s = (i, j + 1)
t = (i, j - 1)
u = (i - 1, j - 1)
v = (i - 1, j)
w = (i - 1, j + 1)
x = (i + 1, j - 1)
y = (i + 1, j)
z = (i + 1, j + 1)
tempholding = [s, t, u, v, w, x, y, z]
for a in tempholding:
if a[0]>=0 and a[1]>=0 and a[0]<=len(grid)-1 and a[1]<=len(grid[i])-1 and grid[a[0]][a[1]]==0:
if (a[0]!=i and a[1]!=j) and (a not in list_to_hold_tuples):
find_connected_0_value_cells(a[0],a[1])
list_to_hold_tuples.append(a)
elif a[0]>=0 and a[1]>=0 and a[0]<=len(grid)-1 and a[1]<=len(grid[i])-1:
list_to_hold_tuples.append(a) #this part adds surrounding non-zero numbers
def reveal_empty_cells():
global booleangrid
for a,b in list_to_hold_tuples:
if a>=0 and a<len(booleangrid) and b>=0 and b<len(booleangrid[a]):
booleangrid[a][b] = False
for i in booleangrid:
print(i)
find_connected_0_value_cells(5,8)#Just using coordinates of 5,8 for testing, this would be click spot
reveal_empty_cells()
我已经重构并让其余部分正常工作。 谢谢@zettatekdev
grid = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 1, 2, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 2, 2, 0, 0, 0, 0, 0, 0],
[1, 1, 2, 3, 0, 0, 0, 0, 0, 0],
[1, 1, 2, 5, 0, 0, 0, 0, 0, 0],
[1, 1, 2, 5, 0, 0, 0, 0, 0, 0]]
list_to_hold_tuples = []
list_change_boolgrid =[]
row = 6
cell = 10
booleangrid = [[True for x in range(cell)] for x in range(row)]
def find_connected_0_value_cells(a, b):
list_to_hold_tuples.append((a, b))
if grid[a][b] == 0:
coord_list = get_surrounding_coords(a, b)
for a,b in coord_list:
if check_coord_values(a, b):
if grid[a][b] != 0:
c = a,b
if c not in list_to_hold_tuples:
list_to_hold_tuples.append(c)
else:
c = a,b
if c not in list_to_hold_tuples:
find_connected_0_value_cells(a,b)
def add_surrounding_cells():
extra_coord = True
for a,b in list_to_hold_tuples:
if grid[a][b] == 0:
coord_list = get_surrounding_coords(a,b, extra_coord)
for i in coord_list:
if i not in list_change_boolgrid:
list_change_boolgrid.append(i)
else:
c = a,b
if c not in list_change_boolgrid:
list_change_boolgrid.append(c)
def reveal_empty_cells():
global booleangrid
for a, b in list_change_boolgrid:
if check_coord_values(a,b):
booleangrid[a][b] = False
def check_coord_values(a,b):
if a == -1 or a >= row:
return False
if b == -1 or b >= cell:
return False
else:
return True
def get_surrounding_coords(a, b, *extra_coord):
c = (a, b + 1)
d = (a, b - 1)
e = (a - 1, b - 1)
f = (a - 1, b)
g = (a - 1, b + 1)
h = (a + 1, b - 1)
i = (a + 1, b)
j = (a + 1, b + 1)
if extra_coord:
k = (a, b)
return [c, d, e, f, g, h, i, j, k]
return [c, d, e, f, g, h, i, j]
find_connected_0_value_cells(3,5)
add_surrounding_cells()
reveal_empty_cells()
print(booleangrid)