连连看有点费脑力，于是我直接用Python写了个自动过关脚本！太爽了！
最近女朋友在玩连连看，玩了一个星期了还没通关，真的是菜。
我实在是看不过去了，直接用python写了个脚本代码，一分钟一把游戏。
快是快，就是联网玩容易被骂，嘿嘿~
模块导入
import cv2 import numpy as np import win32api import win32gui import win32con from PIL import ImageGrab import time import random
窗体标题 用于定位游戏窗体
WINDOW_TITLE = "连连看" # Python学习交流群 815624229 # 本项目素材也在群里可以获取
时间间隔随机生成 [MIN,MAX]
TIME_INTERVAL_MAX = 0.06
TIME_INTERVAL_MIN = 0.1
游戏区域距离顶点的x偏移
MARGIN_LEFT = 10
游戏区域距离顶点的y偏移
MARGIN_HEIGHT = 180
横向的方块数量
H_NUM = 19
纵向的方块数量
V_NUM = 11
方块宽度
POINT_WIDTH = 31
方块高度
POINT_HEIGHT = 35
空图像编号
EMPTY_ID = 0
切片处理时候的左上、右下坐标：
SUB_LT_X = 8 SUB_LT_Y = 8 SUB_RB_X = 27 SUB_RB_Y = 27
游戏的最多消除次数
MAX_ROUND = 200
获取窗体坐标位置
def getGameWindow(): # FindWindow(lpClassName=None, lpWindowName=None) 窗口类名 窗口标题名 window = win32gui.FindWindow(None, WINDOW_TITLE) # 没有定位到游戏窗体 while not window: print('Failed to locate the game window , reposition the game window after 10 seconds...') time.sleep(10) window = win32gui.FindWindow(None, WINDOW_TITLE) # 定位到游戏窗体 # 置顶游戏窗口 win32gui.SetForegroundWindow(window) pos = win32gui.GetWindowRect(window) print("Game windows at " + str(pos)) return (pos[0], pos[1])
获取屏幕截图
def getScreenImage(): print('Shot screen...') # 获取屏幕截图 Image类型对象 scim = ImageGrab.grab() scim.save('screen.png') # 用opencv读取屏幕截图 # 获取ndarray return cv2.imread("screen.png")
从截图中分辨图片 处理成地图
def getAllSquare(screen_image, game_pos): print('Processing pictures...') # 通过游戏窗体定位 # 加上偏移量获取游戏区域 game_x = game_pos[0] + MARGIN_LEFT game_y = game_pos[1] + MARGIN_HEIGHT # 从游戏区域左上开始 # 把图像按照具体大小切割成相同的小块 # 切割标准是按照小块的横纵坐标 all_square = [] for x in range(0, H_NUM): for y in range(0, V_NUM): # ndarray的切片方法 ： [纵坐标起始位置：纵坐标结束为止，横坐标起始位置：横坐标结束位置] square = screen_image[game_y + y * POINT_HEIGHT:game_y + (y + 1) * POINT_HEIGHT, game_x + x * POINT_WIDTH:game_x + (x + 1) * POINT_WIDTH] all_square.append(square) # 因为有些图片的边缘会造成干扰，所以统一把图片往内缩小一圈 # 对所有的方块进行处理 ，去掉边缘一圈后返回 finalresult = [] for square in all_square: s = square[SUB_LT_Y:SUB_RB_Y, SUB_LT_X:SUB_RB_X] finalresult.append(s) return finalresult
判断列表中是否存在相同图形
存在返回进行判断图片所在的id
否则返回-1
def isImageExist(img, img_list): i = 0 for existed_img in img_list: # 两个图片进行比较 返回的是两个图片的标准差 b = np.subtract(existed_img, img) # 若标准差全为0 即两张图片没有区别 if not np.any(b): return i i = i + 1 return -1
获取所有的方块类型
def getAllSquareTypes(all_square): print("Init pictures types...") types = [] # number列表用来记录每个id的出现次数 number = [] # 当前出现次数最多的方块 # 这里我们默认出现最多的方块应该是空白块 nowid = 0; for square in all_square: nid = isImageExist(square, types) # 如果这个图像不存在则插入列表 if nid == -1: types.append(square) number.append(1); else: # 若这个图像存在则给计数器 + 1 number[nid] = number[nid] + 1 if (number[nid] > number[nowid]): nowid = nid # 更新EMPTY_ID # 即判断在当前这张图中的空白块id global EMPTY_ID EMPTY_ID = nowid print('EMPTY_ID = ' + str(EMPTY_ID)) return types
将二维图片矩阵转换为二维数字矩阵
注意因为在上面对截屏切片时是以列为优先切片的
所以生成的record二维矩阵每行存放的其实是游戏屏幕中每列的编号
换个说法就是record其实是游戏屏幕中心对称后的列表
def getAllSquareRecord(all_square_list, types): print("Change map...") record = [] line = [] for square in all_square_list: num = 0 for type in types: res = cv2.subtract(square, type) if not np.any(res): line.append(num) break num += 1 # 每列的数量为V_NUM # 那么当当前的line列表中存在V_NUM个方块时我们认为本列处理完毕 if len(line) == V_NUM: print(line); record.append(line) line = [] return record
判断给出的两个图像能否消除
def canConnect(x1, y1, x2, y2, r): result = r[:] # 如果两个图像中有一个为0 直接返回False if result[x1][y1] == EMPTY_ID or result[x2][y2] == EMPTY_ID: return False if x1 == x2 and y1 == y2: return False if result[x1][y1] != result[x2][y2]: return False # 判断横向连通 if horizontalCheck(x1, y1, x2, y2, result): return True # 判断纵向连通 if verticalCheck(x1, y1, x2, y2, result): return True # 判断一个拐点可连通 if turnOnceCheck(x1, y1, x2, y2, result): return True # 判断两个拐点可连通 if turnTwiceCheck(x1, y1, x2, y2, result): return True # 不可联通返回False return False
判断横向联通
def horizontalCheck(x1, y1, x2, y2, result): if x1 == x2 and y1 == y2: return False if x1 != x2: return False startY = min(y1, y2) endY = max(y1, y2) # 判断两个方块是否相邻 if (endY - startY) == 1: return True # 判断两个方块通路上是否都是0，有一个不是，就说明不能联通，返回false for i in range(startY + 1, endY): if result[x1][i] != EMPTY_ID: return False return True
判断纵向联通
def verticalCheck(x1, y1, x2, y2, result): if x1 == x2 and y1 == y2: return False if y1 != y2: return False startX = min(x1, x2) endX = max(x1, x2) # 判断两个方块是否相邻 if (endX - startX) == 1: return True # 判断两方块儿通路上是否可连。 for i in range(startX + 1, endX): if result[i][y1] != EMPTY_ID: return False return True
判断一个拐点可联通
def turnOnceCheck(x1, y1, x2, y2, result): if x1 == x2 or y1 == y2: return False cx = x1 cy = y2 dx = x2 dy = y1 # 拐点为空，从第一个点到拐点并且从拐点到第二个点可通，则整条路可通。 if result[cx][cy] == EMPTY_ID: if horizontalCheck(x1, y1, cx, cy, result) and verticalCheck(cx, cy, x2, y2, result): return True if result[dx][dy] == EMPTY_ID: if verticalCheck(x1, y1, dx, dy, result) and horizontalCheck(dx, dy, x2, y2, result): return True return False
判断两个拐点可联通
def turnTwiceCheck(x1, y1, x2, y2, result): if x1 == x2 and y1 == y2: return False # 遍历整个数组找合适的拐点 for i in range(0, len(result)): for j in range(0, len(result[1])): # 不为空不能作为拐点 if result[i][j] != EMPTY_ID: continue # 不和被选方块在同一行列的不能作为拐点 if i != x1 and i != x2 and j != y1 and j != y2: continue # 作为交点的方块不能作为拐点 if (i == x1 and j == y2) or (i == x2 and j == y1): continue if turnOnceCheck(x1, y1, i, j, result) and ( horizontalCheck(i, j, x2, y2, result) or verticalCheck(i, j, x2, y2, result)): return True if turnOnceCheck(i, j, x2, y2, result) and ( horizontalCheck(x1, y1, i, j, result) or verticalCheck(x1, y1, i, j, result)): return True return False
自动消除
def autoRelease(result, game_x, game_y): # 遍历地图 for i in range(0, len(result)): for j in range(0, len(result[0])): # 当前位置非空 if result[i][j] != EMPTY_ID: # 再次遍历地图 寻找另一个满足条件的图片 for m in range(0, len(result)): for n in range(0, len(result[0])): if result[m][n] != EMPTY_ID: # 若可以执行消除 if canConnect(i, j, m, n, result): # 消除的两个位置设置为空 result[i][j] = EMPTY_ID result[m][n] = EMPTY_ID print('Remove ：' + str(i + 1) + ',' + str(j + 1) + ' and ' + str(m + 1) + ',' + str( n + 1)) # 计算当前两个位置的图片在游戏中应该存在的位置 x1 = game_x + j * POINT_WIDTH y1 = game_y + i * POINT_HEIGHT x2 = game_x + n * POINT_WIDTH y2 = game_y + m * POINT_HEIGHT # 模拟鼠标点击第一个图片所在的位置 win32api.SetCursorPos((x1 + 15, y1 + 18)) win32api.mouse_event(win32con.MOUSEEVENTF_LEFTDOWN, x1 + 15, y1 + 18, 0, 0) win32api.mouse_event(win32con.MOUSEEVENTF_LEFTUP, x1 + 15, y1 + 18, 0, 0) # 等待随机时间 ，防止检测 time.sleep(random.uniform(TIME_INTERVAL_MIN, TIME_INTERVAL_MAX)) # 模拟鼠标点击第二个图片所在的位置 win32api.SetCursorPos((x2 + 15, y2 + 18)) win32api.mouse_event(win32con.MOUSEEVENTF_LEFTDOWN, x2 + 15, y2 + 18, 0, 0) win32api.mouse_event(win32con.MOUSEEVENTF_LEFTUP, x2 + 15, y2 + 18, 0, 0) time.sleep(random.uniform(TIME_INTERVAL_MIN, TIME_INTERVAL_MAX)) # 执行消除后返回True return True return False
效果的话得上传视频，截图展现不出来效果，大家可以自行试试。
# -*- coding:utf-8 -*- import cv2 import numpy as np import win32api import win32gui import win32con from PIL import ImageGrab import time import random # 窗体标题 用于定位游戏窗体 WINDOW_TITLE = "连连看" # 时间间隔随机生成 [MIN,MAX] TIME_INTERVAL_MAX = 0.06 TIME_INTERVAL_MIN = 0.1 # 游戏区域距离顶点的x偏移 MARGIN_LEFT = 10 # 游戏区域距离顶点的y偏移 MARGIN_HEIGHT = 180 # 横向的方块数量 H_NUM = 19 # 纵向的方块数量 V_NUM = 11 # 方块宽度 POINT_WIDTH = 31 # 方块高度 POINT_HEIGHT = 35 # 空图像编号 EMPTY_ID = 0 # 切片处理时候的左上、右下坐标： SUB_LT_X = 8 SUB_LT_Y = 8 SUB_RB_X = 27 SUB_RB_Y = 27 # 游戏的最多消除次数 MAX_ROUND = 200 def getGameWindow(): # FindWindow(lpClassName=None, lpWindowName=None) 窗口类名 窗口标题名 window = win32gui.FindWindow(None, WINDOW_TITLE) # 没有定位到游戏窗体 while not window: print('Failed to locate the game window , reposition the game window after 10 seconds...') time.sleep(10) window = win32gui.FindWindow(None, WINDOW_TITLE) # 定位到游戏窗体 # 置顶游戏窗口 win32gui.SetForegroundWindow(window) pos = win32gui.GetWindowRect(window) print("Game windows at " + str(pos)) return (pos[0], pos[1]) def getScreenImage(): print('Shot screen...') # 获取屏幕截图 Image类型对象 scim = ImageGrab.grab() scim.save('screen.png') # 用opencv读取屏幕截图 # 获取ndarray return cv2.imread("screen.png") def getAllSquare(screen_image, game_pos): print('Processing pictures...') # 通过游戏窗体定位 # 加上偏移量获取游戏区域 game_x = game_pos[0] + MARGIN_LEFT game_y = game_pos[1] + MARGIN_HEIGHT # 从游戏区域左上开始 # 把图像按照具体大小切割成相同的小块 # 切割标准是按照小块的横纵坐标 all_square = [] for x in range(0, H_NUM): for y in range(0, V_NUM): # ndarray的切片方法 ： [纵坐标起始位置：纵坐标结束为止，横坐标起始位置：横坐标结束位置] square = screen_image[game_y + y * POINT_HEIGHT:game_y + (y + 1) * POINT_HEIGHT, game_x + x * POINT_WIDTH:game_x + (x + 1) * POINT_WIDTH] all_square.append(square) # 因为有些图片的边缘会造成干扰，所以统一把图片往内缩小一圈 # 对所有的方块进行处理 ，去掉边缘一圈后返回 finalresult = [] for square in all_square: s = square[SUB_LT_Y:SUB_RB_Y, SUB_LT_X:SUB_RB_X] finalresult.append(s) return finalresult # 判断列表中是否存在相同图形 # 存在返回进行判断图片所在的id # 否则返回-1 def isImageExist(img, img_list): i = 0 for existed_img in img_list: # 两个图片进行比较 返回的是两个图片的标准差 b = np.subtract(existed_img, img) # 若标准差全为0 即两张图片没有区别 if not np.any(b): return i i = i + 1 return -1 def getAllSquareTypes(all_square): print("Init pictures types...") types = [] # number列表用来记录每个id的出现次数 number = [] # 当前出现次数最多的方块 # 这里我们默认出现最多的方块应该是空白块 nowid = 0; for square in all_square: nid = isImageExist(square, types) # 如果这个图像不存在则插入列表 if nid == -1: types.append(square) number.append(1); else: # 若这个图像存在则给计数器 + 1 number[nid] = number[nid] + 1 if (number[nid] > number[nowid]): nowid = nid # 更新EMPTY_ID # 即判断在当前这张图中的空白块id global EMPTY_ID EMPTY_ID = nowid print('EMPTY_ID = ' + str(EMPTY_ID)) return types # 将二维图片矩阵转换为二维数字矩阵 # 注意因为在上面对截屏切片时是以列为优先切片的 # 所以生成的record二维矩阵每行存放的其实是游戏屏幕中每列的编号 # 换个说法就是record其实是游戏屏幕中心对称后的列表 def getAllSquareRecord(all_square_list, types): print("Change map...") record = [] line = [] for square in all_square_list: num = 0 for type in types: res = cv2.subtract(square, type) if not np.any(res): line.append(num) break num += 1 # 每列的数量为V_NUM # 那么当当前的line列表中存在V_NUM个方块时我们认为本列处理完毕 if len(line) == V_NUM: print(line); record.append(line) line = [] return record def canConnect(x1, y1, x2, y2, r): result = r[:] # 如果两个图像中有一个为0 直接返回False if result[x1][y1] == EMPTY_ID or result[x2][y2] == EMPTY_ID: return False if x1 == x2 and y1 == y2: return False if result[x1][y1] != result[x2][y2]: return False # 判断横向连通 if horizontalCheck(x1, y1, x2, y2, result): return True # 判断纵向连通 if verticalCheck(x1, y1, x2, y2, result): return True # 判断一个拐点可连通 if turnOnceCheck(x1, y1, x2, y2, result): return True # 判断两个拐点可连通 if turnTwiceCheck(x1, y1, x2, y2, result): return True # 不可联通返回False return False def horizontalCheck(x1, y1, x2, y2, result): if x1 == x2 and y1 == y2: return False if x1 != x2: return False startY = min(y1, y2) endY = max(y1, y2) # 判断两个方块是否相邻 if (endY - startY) == 1: return True # 判断两个方块通路上是否都是0，有一个不是，就说明不能联通，返回false for i in range(startY + 1, endY): if result[x1][i] != EMPTY_ID: return False return True def verticalCheck(x1, y1, x2, y2, result): if x1 == x2 and y1 == y2: return False if y1 != y2: return False startX = min(x1, x2) endX = max(x1, x2) # 判断两个方块是否相邻 if (endX - startX) == 1: return True # 判断两方块儿通路上是否可连。 for i in range(startX + 1, endX): if result[i][y1] != EMPTY_ID: return False return True def turnOnceCheck(x1, y1, x2, y2, result): if x1 == x2 or y1 == y2: return False cx = x1 cy = y2 dx = x2 dy = y1 # 拐点为空，从第一个点到拐点并且从拐点到第二个点可通，则整条路可通。 if result[cx][cy] == EMPTY_ID: if horizontalCheck(x1, y1, cx, cy, result) and verticalCheck(cx, cy, x2, y2, result): return True if result[dx][dy] == EMPTY_ID: if verticalCheck(x1, y1, dx, dy, result) and horizontalCheck(dx, dy, x2, y2, result): return True return False def turnTwiceCheck(x1, y1, x2, y2, result): if x1 == x2 and y1 == y2: return False # 遍历整个数组找合适的拐点 for i in range(0, len(result)): for j in range(0, len(result[1])): # 不为空不能作为拐点 if result[i][j] != EMPTY_ID: continue # 不和被选方块在同一行列的不能作为拐点 if i != x1 and i != x2 and j != y1 and j != y2: continue # 作为交点的方块不能作为拐点 if (i == x1 and j == y2) or (i == x2 and j == y1): continue if turnOnceCheck(x1, y1, i, j, result) and ( horizontalCheck(i, j, x2, y2, result) or verticalCheck(i, j, x2, y2, result)): return True if turnOnceCheck(i, j, x2, y2, result) and ( horizontalCheck(x1, y1, i, j, result) or verticalCheck(x1, y1, i, j, result)): return True return False def autoRelease(result, game_x, game_y): # 遍历地图 for i in range(0, len(result)): for j in range(0, len(result[0])): # 当前位置非空 if result[i][j] != EMPTY_ID: # 再次遍历地图 寻找另一个满足条件的图片 for m in range(0, len(result)): for n in range(0, len(result[0])): if result[m][n] != EMPTY_ID: # 若可以执行消除 if canConnect(i, j, m, n, result): # 消除的两个位置设置为空 result[i][j] = EMPTY_ID result[m][n] = EMPTY_ID print('Remove ：' + str(i + 1) + ',' + str(j + 1) + ' and ' + str(m + 1) + ',' + str( n + 1)) # 计算当前两个位置的图片在游戏中应该存在的位置 x1 = game_x + j * POINT_WIDTH y1 = game_y + i * POINT_HEIGHT x2 = game_x + n * POINT_WIDTH y2 = game_y + m * POINT_HEIGHT # 模拟鼠标点击第一个图片所在的位置 win32api.SetCursorPos((x1 + 15, y1 + 18)) win32api.mouse_event(win32con.MOUSEEVENTF_LEFTDOWN, x1 + 15, y1 + 18, 0, 0) win32api.mouse_event(win32con.MOUSEEVENTF_LEFTUP, x1 + 15, y1 + 18, 0, 0) # 等待随机时间 ，防止检测 time.sleep(random.uniform(TIME_INTERVAL_MIN, TIME_INTERVAL_MAX)) # 模拟鼠标点击第二个图片所在的位置 win32api.SetCursorPos((x2 + 15, y2 + 18)) win32api.mouse_event(win32con.MOUSEEVENTF_LEFTDOWN, x2 + 15, y2 + 18, 0, 0) win32api.mouse_event(win32con.MOUSEEVENTF_LEFTUP, x2 + 15, y2 + 18, 0, 0) time.sleep(random.uniform(TIME_INTERVAL_MIN, TIME_INTERVAL_MAX)) # 执行消除后返回True return True return False def autoRemove(squares, game_pos): game_x = game_pos[0] + MARGIN_LEFT game_y = game_pos[1] + MARGIN_HEIGHT # 重复一次消除直到到达最多消除次数 while True: if not autoRelease(squares, game_x, game_y): # 当不再有可消除的方块时结束 ， 返回消除数量 return if __name__ == '__main__': random.seed() # i. 定位游戏窗体 game_pos = getGameWindow() time.sleep(1) # ii. 获取屏幕截图 screen_image = getScreenImage() # iii. 对截图切片，形成一张二维地图 all_square_list = getAllSquare(screen_image, game_pos) # iv. 获取所有类型的图形，并编号 types = getAllSquareTypes(all_square_list) # v. 讲获取的图片地图转换成数字矩阵 result = np.transpose(getAllSquareRecord(all_square_list, types)) # vi. 执行消除 , 并输出消除数量 print('The total elimination amount is ' + str(autoRemove(result, game_pos)))
兄弟们快去试试吧
