八皇后问题 -- python面向对象解法
2024-08-27 14:29:05
# 【8*8棋盘八皇后问题】 class Queen:
def __init__(self, row, col):
self.row = row
self.col = col
self.pos = (row + 1, col + 1) def check(self, chess_board):
'''检查当前位置是否可放置皇后,默认每行只放一个皇后'''
# 检测列
for i in range(8):
if chess_board.board[i][self.col] == 0:
return False
# 检测两个斜向
for i in range(-7, 8):
if 0 <= self.row + i < 8 and 0 <= self.col + i < 8 and chess_board.board[self.row + i][self.col + i] == 0:
return False
elif 0 <= self.row + i < 8 and 0 <= self.col - i < 8 and chess_board.board[self.row + i][self.col - i] == 0:
return False
return True class Chess_Board:
# 棋盘 无皇后为1,有皇后为0
board = [[1 for i in range(8)] for i in range(8)] def print_board(self):
'''可视化打印棋盘'''
print('---------------------------------')
for i in range(8):
for j in range(8):
if self.board[i][j] == 1:
print(' +', end='') # + 无皇后
else:
print(' @', end='') # @ 表示皇后
print()
print('---------------------------------') def solve():
'''返回结果棋盘对象和皇后对象列表''' def check_nextline(queens, chess_board):
'''已知所有皇后坐标,判断下一行是否有位置'''
next_col = len(queens)
if next_col < 8:
for i in range(8):
queen = Queen(next_col, i)
if queen.check(chess_board):
queens.append(queen)
chess_board.board[queen.row][queen.col] = 0
# chess_board.print_board() # 监控实现过程
return True
return False queens = [] # 存储已放到棋盘上的皇后对象的列表
chess_board = Chess_Board() # 创建唯一棋盘对象
queen = Queen(0, 0)
queens.append(queen)
chess_board.board[queen.row][queen.col] = 0
while 1:
# 如果下一行没有位置,就将该行的皇后向右移一格
while not check_nextline(queens, chess_board):
queen = queens.pop(-1)
chess_board.board[queen.row][queen.col] = 1
while queen.col == 7:
queen = queens.pop(-1)
chess_board.board[queen.row][queen.col] = 1 queen = Queen(queen.row, queen.col + 1)
# 如果皇后右移后与原先皇后冲突,则继续右移
while not queen.check(chess_board):
if queen.col < 7:
queen = Queen(queen.row, queen.col + 1)
else:
while queen.col == 7:
queen = queens.pop(-1)
chess_board.board[queen.row][queen.col] = 1 queen = Queen(queen.row, queen.col + 1)
queens.append(queen)
chess_board.board[queen.row][queen.col] = 0 # 放满8个皇后之后跳出while 1循环
if len(queens) == 8:
break
return queens, chess_board def main():
'''主函数'''
queens, chess_board = solve()
for queen in queens:
print(queen.pos)
chess_board.print_board() if __name__ == "__main__":
main()
最新文章
- Android探索之Service全面回顾及总结
- [译文]通过ID, TagName, ClassName, Name, CSS selector 得到element
- python 传参open
- 【java】:常用工具类
- 【leetcode】Isomorphic Strings(easy)
- mysql存储过程出参入参,sqlserver很熟悉的一件事到mysql,捣鼓了大半天。记录一下提醒自己。勿看
- html5中的表单
- leetcode 37 Sudoku Solver java
- jQuery 学习笔记(未完待续)
- Android入门教程之我见
- Android 上多方式定位元素(python)
- JBPM4.4 基本使用
- linux 硬盘满了如何处理
- 常用sql语句总结(二)(更新数据,序列,创建数据表,约束,注释)
- mac新手使用
- 从零开始学 Web 之 JavaScript(二)变量
- 玩弄 python 正则表达式
- Vim/Vi的使用
- Mysql表连接查询
- Laravel框架中Validor中错误信息$error的输出