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"""
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@Author: packy945
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@FileName: sudoku.py
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@DateTime: 2023/4/27 17:41
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@SoftWare: PyCharm
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"""
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# -*- encoding: utf-8 -*-
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from data import *
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from UI import *
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def game_menu(menu:tk.Menu):
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"""游戏菜单显示"""
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menu_node = Menu(menu, tearoff=False)
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menu_node.add_command(label="重新开始", command=lambda: restart(N))
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menu_node.add_command(label="暂停计时", command=lambda: stop_time())
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menu_node.add_command(label="载入题目", command=lambda: load_game())
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# 在主目录菜单上新增"菜单"选项,并通过menu参数与下拉菜单绑定
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menu.add_cascade(label="游戏设置", menu=menu_node)
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def auto_menu(menu:tk.Menu):
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"""自动求解菜单显示"""
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menu_node = Menu(menu, tearoff=False)
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menu_node.add_command(label="自动求解-顺序求解", command=lambda: [init_u_M(), auto_Solve(u_martix, STEP.copy())])
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menu_node.add_command(label="自动求解-优化", command=lambda: [init_u_M(), auto_Solve_new(u_martix, STEP.copy())])
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fmenu = Menu(menu, tearoff=False)
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fmenu.add_command(label='低速', command=lambda: speed_set(1))
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fmenu.add_command(label='中速', command=lambda: speed_set(0.5))
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fmenu.add_command(label='高速', command=lambda: speed_set(0.2))
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menu_node.add_cascade(label="求解速度选择", menu=fmenu)
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# 在主目录菜单上新增"菜单"选项,并通过menu参数与下拉菜单绑定
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menu.add_cascade(label="自动求解", menu=menu_node)
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main_menu = tk.Menu(root)
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root.config(menu=main_menu)
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game_menu(main_menu)
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auto_menu(main_menu)
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def speed_set(spd):
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global wait_time
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wait_time = spd
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def stop_time():
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global time1
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time1.stop()
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def init_u_M():
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global martix, u_martix, step, STEP, seq
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u_martix = martix.copy()
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step = [] # 操作步骤
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STEP = [[0, (0, 0), 0, martix]] # 记录每一步的状态
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seq = 1 # 全局操作次序初始化
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def restart(N):
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'''
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重新生成一个游戏
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:return:
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'''
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global martix, u_martix, time1, show_possible, loca_o, loca, note_mark, step, STEP, seq
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time1.restart()
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martix = InitMartix(N) # 生成一个数独数组
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u_martix = martix.copy() # 复制用户martix数组
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show_possible = 0 # 是否显示提示
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p = []# 初始化可能值数组
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possible = []
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for i in range(9):
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p.append({}.copy())
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for i in range(9):
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possible.append(p.copy())
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loca_o = tuple() # martix 数组上次滞留坐标
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loca = tuple()# martix 数组当前坐标
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note_mark = False # 笔记开启标志
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show_Time(time1)# 显示时间信息
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step = [] # 操作步骤
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STEP = [[0, (0, 0), 0, martix]] # 记录每一步的状态
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seq = 1 # 全局操作次序初始化
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draw(loca, martix, u_martix, STEP)
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def auto_Solve(u_M, STEP):
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'''
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自动解题算法(按顺序求解)
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:param u_M:待解答的数独矩阵
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:return:求解完毕的矩阵
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'''
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for row in range(9):
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for col in range(9):
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if u_M[row, col] == 0:
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possible = GetPossible(u_M, row, col) # 计算所有的可能的数字
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for value in possible:#尝试所有可能数字
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u_M[row, col] = value # 将可能的数组填入
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seq = STEP[-1][0] + 1 # 记录顺序
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step = [seq, (row + 1, col + 1), value, u_M.copy()] # 记录一步
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STEP.append(step)#将操作记录加入Step中
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draw((row + 1, col + 1), martix, u_M, STEP)#更新显示界面
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time.sleep(wait_time) # 防止运算过快,减慢演算步骤
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if auto_Solve(u_M, STEP.copy()): # 继续深度优先遍历填入数字
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return True # 求解完成,直接返回
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u_M[row, col] = 0 # 若用这个数无解,则重置为0,进行回溯
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draw((row + 1, col + 1), martix, u_M, STEP)#更新显示界面
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time.sleep(wait_time) # 停顿一段时间,显示求解步骤
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return False #当前无解,返回False
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return True# 当所有的数字填完,数独求解完毕
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def auto_Solve_new(u_M, STEP):
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'''
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自动求解(优化)
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:param u_M:待求解数独矩阵
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:return:求解完成的矩阵
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'''
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flag = 0
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# 优化:如果某个位置只有一个可能值,则先填这个格子
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for row in range(9):
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for col in range(9):
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if u_M[row, col] == 0:
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possible = GetPossible(u_M, row, col) # 所有的可能的数字
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if len(possible) == 1:
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flag = 1#标记可以进行求解优化
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value = possible.pop()#只有一个可能数组,则直接将该数字填入
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u_M[row, col] = value # 将确定的数字填入填入
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seq = STEP[-1][0] + 1 # 记录顺序
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step = [seq, (row + 1, col + 1), value, u_M.copy()] # 记录一步
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STEP.append(step)#将步骤记录加到记录中
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draw((row + 1, col + 1), martix, u_M, STEP) #重新绘图
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time.sleep(wait_time) # 停顿一段时间,显示求解步骤
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if auto_Solve_new(u_M, STEP): # 继续深度优先遍历填入数字
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return True # 填完最后一个数字
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u_M[row, col] = 0 # 如果当前填入的数字会导致后面无解则依然填入0表示空白待填
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draw((row + 1, col + 1), martix, u_M, STEP)#重新绘图
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time.sleep(wait_time) # 停顿一段时间,显示求解步骤
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if not flag:#若没有能直接确定的格子,则按原方法进行求解
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for row in range(9):
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for col in range(9):
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if u_M[row, col] == 0:
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possible = GetPossible(u_M, row, col) # 所有的可能的数字
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for value in possible:
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u_M[row, col] = value # 将可能的数组填入
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seq = STEP[-1][0] + 1 # 记录顺序
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step = [seq, (row + 1, col + 1), value, u_M.copy()] # 记录一步
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STEP.append(step)
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draw((row + 1, col + 1), martix, u_M, STEP)
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time.sleep(wait_time) # 停顿一段时间,显示求解步骤
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if auto_Solve_new(u_M, STEP): # 继续深度优先遍历填入数字
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return True # 填完最后一个数字
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u_M[row, col] = 0 # 如果当前填入的数字会导致后面无解则依然填入0表示空白待填
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draw((row + 1, col + 1), martix, u_M, STEP)
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time.sleep(wait_time) # 停顿一段时间,显示求解步骤
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return False
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return True# 当所有的数字填完,数独求解完毕
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def load_game():
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'''
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载入存储的游戏
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:return:
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'''
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global martix, u_martix, time1, show_possible, loca_o, loca, note_mark, step, STEP, seq
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time1.restart()
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martix = np.load("martix.npy")
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u_martix = martix.copy() # 复制用户martix数组
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show_possible = 0 # 是否显示提示
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# 初始化可能值数组
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p = []
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possible = []
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for i in range(9):
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p.append({}.copy())
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for i in range(9):
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possible.append(p.copy())
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loca_o = tuple() # martix 数组上次滞留坐标
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loca = tuple()
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note_mark = False # 笔记开启标志
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# 显示时间信息
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show_Time(time1)
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step = [] # 操作步骤
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STEP = [[0, (0, 0), 0, martix]] # 记录每一步的状态
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seq = 1 # 全局操作次序初始化
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draw(loca, martix, u_martix, STEP)
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def left1(event):
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'''
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martix坐标映射
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:param event: 鼠标左键点击的坐标值
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:return: 点击的方格坐标
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'''
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""""""
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x = event.x
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y = event.y
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row = -1
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col = -1
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global STEP, loca, loca_o
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for cur_col in range(9):
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for cur_row in range(9):
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top_x, top_y = cur_col * (bc + 3) + 50, cur_row * (bc + 3) + 50# 计算每个方格的四个角
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ud_x, ud_y = (cur_col + 1) * (bc + 3) + 50, (cur_row + 1) * (bc + 3) + 50# 计算每个方格的四个角
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if x < ud_x and x > top_x:#计算点击的列数
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col = cur_col
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if y < ud_y and y > top_y:#计算点击的行数
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row = cur_row
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if row >= 0 and col >= 0 and not martix[row, col]:
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if loca:#若已经点击过坐标
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loca_o = loca # 保留上次点击过的坐标
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loca = (row + 1, col + 1) # 更新选中的格子位置
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draw(loca, martix, u_martix, STEP)#绘制界面
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return
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def win(u_martix, timer: MyTimer, STEP):
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'''
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检查游戏是否结束,若结束则输出结束界面
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:param u_martix:用户矩阵
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:param timer:计时器
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:return:
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'''
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count = 0 # 计数
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for i in range(9):
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for j in range(9):
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if u_martix[i][j] == 0 or u_martix[i][j] == 10:
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count += 1
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if count == 0:
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if Judge(u_martix):
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for w in right.winfo_children():
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w.destroy()
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timer.stop()
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win_words = tk.Label(right, text=f"游戏结束,恭喜通过", bg='white', fg='red', font=('幼圆', 16, 'bold'),
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anchor=W)
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win_words.place(relx=0.1, rely=0.3)
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win_time = tk.Label(right, text=f"耗时:{timer.min.get()}分{timer.sec.get()}秒", bg='white', fg='red',
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font=('幼圆', 16, 'bold'), anchor=W)
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win_time.place(relx=0.1, rely=0.4)
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win_step = tk.Label(right, text=f"花费步数:{len(STEP)}步", bg='white', fg='red',
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font=('幼圆', 16, 'bold'), anchor=W)
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win_step.place(relx=0.1, rely=0.5)
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re = tk.Button(right, text=f"重新开始", bg='white', fg='black',
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font=('幼圆', 14, 'bold'), command=lambda: restart(N))
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re.place(relx=0.1, rely=0.6, relwidth=0.5)
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def u_number(num, loca):
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'''
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添加功能:
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:param num: 用户点击数字按钮
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:param loca: 当前位置
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:return:
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'''
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if loca != ():#若选择了一个格子
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row, col = loca[0] - 1, loca[1] - 1
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global step, seq
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if num: # 直接点击了数字按钮,就添加num
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if martix[row][col] == 0 and not note_mark: # 空格且笔记模式为False才可以填写
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if u_martix[row][col] == 10: # 如果修改了这个格子的值,删除原来的记录
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possible[row][col].clear()
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# for note in note_martix: # 如果这个空格有笔记直接覆盖清空笔记
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# if note[0] == row and note[1] == col:
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# note_martix.remove(note)
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u_martix[row][col] = num # 为用户数组添加数值
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seq = STEP[-1][0] + 1 # 记录顺序
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step = [seq, loca, num, u_martix.copy()] # 记录一步
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# print(step[:3])
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if len(step) != 0: # 添加步骤
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STEP.append(step)
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step = []
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|
|
|
|
print(loca, num, "添加")
|
|
|
|
|
elif note_mark: # 笔记模式打开时
|
|
|
|
|
if u_martix[row][col] == 0:# 若当前没有笔记,则只有num
|
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|
|
|
u_martix[row][col] = 10#标记该格为笔记格子
|
|
|
|
|
possible[row][col] = {num}# 将笔记加入记录中
|
|
|
|
|
elif u_martix[row][col] == 10:# 若已有笔记,则添加num
|
|
|
|
|
possible[row][col].add(num)
|
|
|
|
|
draw(loca, martix, u_martix, STEP)#重新绘制界面
|
|
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|
|
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|
|
def control_draw_old(bc):
|
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|
|
|
'''
|
|
|
|
|
绘制数字按钮
|
|
|
|
|
:param bc: 全局按钮大小
|
|
|
|
|
:return:
|
|
|
|
|
'''
|
|
|
|
|
for i in range(9):#遍历0~9
|
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|
|
top_x, top_y = i * (bc + 3) + 50, 12 * (bc + 3) + 10#计算每个按钮左上角坐标
|
|
|
|
|
B = tk.Button(root, text=f"{i + 1}", bg='white', fg='black',
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|
|
font=('幼圆', 14, 'bold'), command=lambda num=i + 1: u_number(num, loca))
|
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|
B.place(x=top_x, y=top_y, width=bc + 3, height=bc + 3)#绘制按钮到图上
|
|
|
|
|
def control_draw(bc, u_martix):
|
|
|
|
|
'''
|
|
|
|
|
绘制数字按钮
|
|
|
|
|
:param bc: 全局按钮大小
|
|
|
|
|
:param u_martix: 当前用户矩阵
|
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|
|
|
:return:
|
|
|
|
|
'''
|
|
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|
|
num = [9] * 11#记录每个数字缺失的个数为9
|
|
|
|
|
for i in range(9):
|
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|
|
for j in range(9):#遍历当前数组,每有一个数字,则该数字的缺失值减一
|
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|
|
num[u_martix[i][j]] = num[u_martix[i][j]] - 1
|
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|
|
for i in range(9):
|
|
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|
|
top_x, top_y = i * (bc + 3) + 50, 12 * (bc + 3) + 10#计算每个按钮左上角坐标
|
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|
|
|
if num[i + 1]:#如果该数字还有缺失(图中不到9个),则按钮可以显示
|
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|
|
B = tk.Button(root, text=f"{i + 1}", bg='white', fg='black',
|
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|
|
font=('幼圆', 14, 'bold'), command=lambda num=i + 1: u_number(num, loca))
|
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|
|
else:#若某个数字已经有9个,说明已经不能选择该数字,设置这个按钮为不可按
|
|
|
|
|
B = tk.Button(root, text=f"{i + 1}", bg='grey', fg='black', state=DISABLED,
|
|
|
|
|
font=('幼圆', 14, 'bold'), command=lambda num=i + 1: u_number(num, loca))
|
|
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|
|
B.place(x=top_x, y=top_y, width=bc + 3, height=bc + 3)#绘制按钮到图上
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
def listen_del(loca, possible):
|
|
|
|
|
'''
|
|
|
|
|
擦除功能
|
|
|
|
|
:param loca: 当前位置
|
|
|
|
|
:return:
|
|
|
|
|
'''
|
|
|
|
|
if loca != ():
|
|
|
|
|
# 选定了位置才能擦除
|
|
|
|
|
row, col = loca[0] - 1, loca[1] - 1
|
|
|
|
|
if not u_martix[row][col] == 0 and martix[row][col] == 0: # 原题目中没有数字才可以删除
|
|
|
|
|
u_martix[row][col] = 0#设置为空
|
|
|
|
|
possible[row][col].clear()#清除可能值
|
|
|
|
|
print(loca, 2, "删除")#输出删除信息
|
|
|
|
|
draw(loca, martix, u_martix, STEP)#显示界面
|
|
|
|
|
|
|
|
|
|
def listen_all_note(possible, u_martix):
|
|
|
|
|
'''
|
|
|
|
|
一键提示功能
|
|
|
|
|
:param pos:
|
|
|
|
|
:return:
|
|
|
|
|
'''
|
|
|
|
|
print('all note')#输出提示信息
|
|
|
|
|
global show_possible, loca
|
|
|
|
|
show_possible = 1 # 显示提示信息
|
|
|
|
|
for row in range(9):
|
|
|
|
|
for col in range(9):
|
|
|
|
|
if u_martix[row][col] == 10 or u_martix[row][col] == 0:# 只有空格才会更新提示
|
|
|
|
|
possible[row][col] = GetPossible(u_martix, row, col)#计算可能取值
|
|
|
|
|
u_martix[row][col] = 10#标记这个格子为存在提示
|
|
|
|
|
draw(loca, martix, u_martix, STEP)#重新绘制界面
|
|
|
|
|
|
|
|
|
|
def show_hint(bc, u_martix, possible):
|
|
|
|
|
'''
|
|
|
|
|
显示提示信息
|
|
|
|
|
:param bc: 方格大小
|
|
|
|
|
:return:
|
|
|
|
|
'''
|
|
|
|
|
hint_bc = int(bc // 3)# 小数字的间距
|
|
|
|
|
for row in range(9):
|
|
|
|
|
for col in range(9):
|
|
|
|
|
if not u_martix[row][col] == 10:
|
|
|
|
|
possible[row][col].clear()# 只有u_martix[j][i]为10才表示该格子有笔记存在
|
|
|
|
|
continue# 若该格无笔记,则跳过该格
|
|
|
|
|
x, y = col * (bc + 3) + 50 + bc / 2, row * (bc + 3) + 50 + bc / 2# 计算大方格中点坐标
|
|
|
|
|
x = x - hint_bc#计算左上角第一个数字坐标
|
|
|
|
|
y = y - hint_bc#计算左上角第一个数字坐标
|
|
|
|
|
for ii in range(3):# 按照九宫格显示小数字
|
|
|
|
|
for jj in range(3):
|
|
|
|
|
n = ii * 3 + jj + 1#遍历1~9
|
|
|
|
|
if possible[row][col]: # 在用户已填的部分有可能导致部分格子无解,所以先加一个判断
|
|
|
|
|
for k in possible[row][col]:
|
|
|
|
|
if n == k:# 如果possible[row][col]中含有k,则说明该格子可以填k,则将k显示在方格中
|
|
|
|
|
cv.create_text(x + ii * hint_bc, y + jj * hint_bc,
|
|
|
|
|
text=f"{n}", font=('楷体', 10), fill='purple')
|
|
|
|
|
cv.update()#更新画布
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
def listen_hint(loca, u_martix):
|
|
|
|
|
'''
|
|
|
|
|
提示功能:显示当前格子的可能值
|
|
|
|
|
:param loca: 当前格子
|
|
|
|
|
:return:
|
|
|
|
|
'''
|
|
|
|
|
if loca != ():
|
|
|
|
|
row, col = loca[0] - 1, loca[1] - 1
|
|
|
|
|
if u_martix[row][col] == 0 or u_martix[row][col] == 10:#只有该方格没有数字时才能提示
|
|
|
|
|
possible[row][col] = GetPossible(u_martix, row, col)#计算该方格可能取值
|
|
|
|
|
u_martix[row][col] = 10#标记该方格为存在提示
|
|
|
|
|
draw(loca, martix, u_martix, STEP)#重新绘制界面
|
|
|
|
|
|
|
|
|
|
def listen_note(loca):
|
|
|
|
|
'''
|
|
|
|
|
笔记功能
|
|
|
|
|
:param loca: 当前位置
|
|
|
|
|
:return:
|
|
|
|
|
'''
|
|
|
|
|
if loca != ():#若已经选中一个方格
|
|
|
|
|
global note_mark # 函数内修改全局变量
|
|
|
|
|
note_mark = not note_mark#切换笔记模式
|
|
|
|
|
print(f"笔记功能:{note_mark}")#输出信息
|
|
|
|
|
draw(loca, martix, u_martix, STEP)#绘制界面
|
|
|
|
|
|
|
|
|
|
def listen_backspace():
|
|
|
|
|
'''
|
|
|
|
|
撤回功能
|
|
|
|
|
:return:
|
|
|
|
|
'''
|
|
|
|
|
global u_martix, STEP
|
|
|
|
|
if len(STEP) > 1:# 如果操作记录中有操作
|
|
|
|
|
STEP.pop()# 删除操作记录最后一步操作
|
|
|
|
|
u_martix = STEP[-1][-1].copy()#将数独状态复位到上一步操作
|
|
|
|
|
loca = STEP[-1][1]#将当前选定方格复位到上一步操作
|
|
|
|
|
draw(loca, martix, u_martix, STEP)#重新绘制界面
|
|
|
|
|
else:#若已经没有操作步骤(已经回到游戏初始状态)
|
|
|
|
|
u_martix = STEP[-1][-1].copy()#将数独状态复制到初始状态
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
def control(bc):
|
|
|
|
|
'''
|
|
|
|
|
绘制五个控制按键
|
|
|
|
|
:param bc:全局按钮大小
|
|
|
|
|
:return:
|
|
|
|
|
'''
|
|
|
|
|
global loca, loca_o
|
|
|
|
|
lon = 9 * (bc + 3) + 50
|
|
|
|
|
lon = lon // 5 - 12 # 控制按键长度
|
|
|
|
|
ctl = ["撤回", "擦除", "笔记", "一键笔记", "提示"]
|
|
|
|
|
# 撤回
|
|
|
|
|
B1 = tk.Button(root, text=f"{ctl[0]}", bg='white', fg='black',
|
|
|
|
|
font=('幼圆', 12, 'bold'), command=lambda: listen_backspace())
|
|
|
|
|
B1.place(x=(lon + 2) * 0 + 50, y=(bc + 3) * 11 + 8, width=lon, height=bc)
|
|
|
|
|
# 擦除
|
|
|
|
|
B2 = tk.Button(root, text=f"{ctl[1]}", bg='white', fg='black',
|
|
|
|
|
font=('幼圆', 12, 'bold'), command=lambda: listen_del(loca, possible))
|
|
|
|
|
B2.place(x=(lon + 2) * 1 + 50, y=(bc + 3) * 11 + 8, width=lon, height=bc)
|
|
|
|
|
# 笔记
|
|
|
|
|
B3 = tk.Button(root, text=f"{ctl[2]}", bg='white', fg='black',
|
|
|
|
|
font=('幼圆', 12, 'bold'), command=lambda: listen_note(loca))
|
|
|
|
|
B3.place(x=(lon + 2) * 2 + 50, y=(bc + 3) * 11 + 8, width=lon, height=bc)
|
|
|
|
|
# 一键笔记
|
|
|
|
|
B4 = tk.Button(root, text=f"{ctl[3]}", bg='white', fg='black',
|
|
|
|
|
font=('幼圆', 12, 'bold'), command=lambda: listen_all_note(possible, u_martix))
|
|
|
|
|
B4.place(x=(lon + 2) * 3 + 50, y=(bc + 3) * 11 + 8, width=lon, height=bc)
|
|
|
|
|
# 提示
|
|
|
|
|
B5 = tk.Button(root, text=f"{ctl[4]}", bg='white', fg='black',
|
|
|
|
|
font=('幼圆', 12, 'bold'), command=lambda: listen_hint(loca, u_martix))
|
|
|
|
|
B5.place(x=(lon + 2) * 4 + 50, y=(bc + 3) * 11 + 8, width=lon, height=bc)
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
def draw(loca, martix, u_martix, STEP):
|
|
|
|
|
'''
|
|
|
|
|
绘制游戏界面
|
|
|
|
|
:param loca: 当前位置
|
|
|
|
|
:return:
|
|
|
|
|
'''
|
|
|
|
|
for i in range(9):
|
|
|
|
|
for j in range(9):# 绘制小方格
|
|
|
|
|
rect_draw(BC, i, j)
|
|
|
|
|
if loca != ():#根据当前选中位置绘制阴影
|
|
|
|
|
row, col = loca[0] - 1, loca[1] - 1
|
|
|
|
|
for i in range(9):
|
|
|
|
|
for j in range(9):
|
|
|
|
|
if col == i and row == j:#当前选中的方格使用灰色标记
|
|
|
|
|
cv.create_rectangle(col * (BC + 3) + 52, row * (BC + 3) + 52, col * (BC + 3) + 48 + BC,
|
|
|
|
|
row * (BC + 3) + 48 + BC, fill='grey', width=0)
|
|
|
|
|
elif col == i or row == j:#当前选中方格所在行列使用浅灰色标记
|
|
|
|
|
cv.create_rectangle(i * (BC + 3) + 52, j * (BC + 3) + 52, i * (BC + 3) + 48 + BC,
|
|
|
|
|
j * (BC + 3) + 48 + BC, fill='#CCCCCC', width=0)
|
|
|
|
|
show_hint(BC, u_martix, possible)# 显示提示信息
|
|
|
|
|
for row in range(9):
|
|
|
|
|
for col in range(9):# 在小方格上绘制数字
|
|
|
|
|
num_draw(BC, row, col, martix, u_martix)
|
|
|
|
|
print(u_martix.tolist())
|
|
|
|
|
print(loca)
|
|
|
|
|
control(BC)# 绘制五个控制按键
|
|
|
|
|
control_draw(BC, u_martix)# 绘制9个数字按钮
|
|
|
|
|
empty_count = empty(martix)# 计算数组空格
|
|
|
|
|
Difficulty(empty_count)# 根据空格数量显示难度
|
|
|
|
|
noter(note_mark)# 显示笔记提示
|
|
|
|
|
seq_recode(right, STEP)# 显示步骤
|
|
|
|
|
win(u_martix, time1, STEP)# 检查游戏是否结束
|
|
|
|
|
cv.update()# 更新画布
|
|
|
|
|
|
|
|
|
|
N = 30 # 空白格子数
|
|
|
|
|
martix = InitMartix(N)
|
|
|
|
|
u_martix = martix.copy() # 复制用户martix数组
|
|
|
|
|
BC = 40
|
|
|
|
|
show_possible = 0 # 是否显示提示
|
|
|
|
|
# 初始化可能值数组
|
|
|
|
|
p = []
|
|
|
|
|
possible = []
|
|
|
|
|
for i in range(9):
|
|
|
|
|
p.append({}.copy())
|
|
|
|
|
for i in range(9):
|
|
|
|
|
possible.append(p.copy())
|
|
|
|
|
# print(possible)
|
|
|
|
|
loca_o = tuple() # martix 数组上次滞留坐标
|
|
|
|
|
loca = tuple()
|
|
|
|
|
note_mark = False # 笔记开启标志
|
|
|
|
|
# note_list = [] # 笔记数字记录
|
|
|
|
|
# note_martix = [] # 笔记位置记录列表
|
|
|
|
|
wait_time = 0.5 # 自动求解速度
|
|
|
|
|
step = [] # 操作步骤
|
|
|
|
|
STEP = [[0, (0, 0), 0, martix]] # 记录每一步的状态
|
|
|
|
|
seq = 1 # 全局操作次序初始化
|
|
|
|
|
|
|
|
|
|
time1 = MyTimer() # 实例计时器
|
|
|
|
|
# 显示时间信息
|
|
|
|
|
show_Time(time1)
|
|
|
|
|
draw(loca, martix, u_martix, STEP)
|
|
|
|
|
|
|
|
|
|
cv.bind('<Button-1>', left1)
|
|
|
|
|
|
|
|
|
|
if __name__ == "__main__":
|
|
|
|
|
mainloop()
|