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~
直接上代码
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"
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。素材可以私我
全部代码
# -*- 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
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