编写一个包含jf的扫雷游戏
时间: 2024-06-09 08:11:31 浏览: 137
以下是一个基本的扫雷游戏,其中包含“JF”功能:
```python
import random
class Minesweeper:
def __init__(self, size, mines):
self.size = size
self.mines = mines
self.board = []
self.mask = []
self.game_over = False
# 初始化棋盘和掩码
for i in range(size):
row = []
mask_row = []
for j in range(size):
row.append(0)
mask_row.append(False)
self.board.append(row)
self.mask.append(mask_row)
# 放置地雷
for i in range(mines):
while True:
x = random.randint(0, size - 1)
y = random.randint(0, size - 1)
if self.board[x][y] != -1:
self.board[x][y] = -1
break
# 计算每个方格周围地雷数量
for i in range(size):
for j in range(size):
if self.board[i][j] == -1:
continue
count = 0
for x in range(max(i-1, 0), min(i+2, size)):
for y in range(max(j-1, 0), min(j+2, size)):
if self.board[x][y] == -1:
count += 1
self.board[i][j] = count
def __str__(self):
result = ""
for i in range(self.size):
for j in range(self.size):
if self.mask[i][j]:
if self.board[i][j] == -1:
result += "X "
elif self.board[i][j] == 0:
result += " "
else:
result += str(self.board[i][j]) + " "
else:
result += "- "
result += "\n"
return result
def reveal(self, x, y):
if self.mask[x][y]:
return False
self.mask[x][y] = True
if self.board[x][y] == -1:
self.game_over = True
return True
if self.board[x][y] == 0:
for i in range(max(x-1, 0), min(x+2, self.size)):
for j in range(max(y-1, 0), min(y+2, self.size)):
self.reveal(i, j)
return False
def mark(self, x, y):
self.mask[x][y] = not self.mask[x][y]
return False
def jf(self):
for i in range(self.size):
for j in range(self.size):
if not self.mask[i][j]:
self.mask[i][j] = True
if self.board[i][j] == -1:
self.game_over = True
return True
return False
# 测试
game = Minesweeper(5, 5)
print(game)
while not game.game_over:
x = int(input("x: "))
y = int(input("y: "))
action = input("action: ")
if action == "r":
if game.reveal(x, y):
print("Game over!")
break
elif action == "m":
game.mark(x, y)
elif action == "jf":
if game.jf():
print("Game over!")
break
print(game)
```
在这个游戏中,玩家可以输入 `jf` 来揭示所有未揭示的方格并找到所有地雷。注意,这个实现中只有一个地雷标志可以被插在一个方格中,所以如果玩家标记了一个错误的方格,游戏会继续进行,但会被认为是失败。
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