本文实例讲述了Python基于分水岭算法解决走迷宫游戏。分享给大家供大家参考，具体如下：

#Solving maze with morphological transformation"""usage:Solving maze with morphological transformationneeded module:cv2/numpy/sysref:1.http:///post/leeyoung/539a629aab35bc44e2000000@author:Robin Chen"""import cv2import numpy as npimport sysdef SolvingMaze(image):#load an image try: img = cv2.imread(image) except Exception,e: print 'Error:can not open the image!' sys.exit()#show image #cv2.namedWindow('image', cv2.WINDOW_NORMAL) cv2.imshow('maze_image',img)#convert to gray gray_image = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)#show gray image #cv2.imshow('gray_image',gray_image)#convert to binary image retval,binary_image = cv2.threshold(gray_image, 10,255, cv2.THRESH_BINARY_INV) #cv2.imshow('binary_image',binary_image) contours,hierarchy = cv2.findContours(binary_image, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_NONE) if len(contours) != 2: sys.exit("This is not a 'perfect maze' with just 2 walls!") h, w, d = img.shape#The first wall path = np.zeros((h,w),dtype = np.uint8)#cv2.CV_8UC1 cv2.drawContours(path, contours, 0, (255,255,255),-1)#cv2.FILLED #cv2.imshow('The first wall',path)#Dilate the wall by a few pixels kernel = np.ones((19, 19), dtype = np.uint8) path = cv2.dilate(path, kernel) #cv2.imshow('Dilate the wall by a few pixels',path)#Erode by the same amount of pixels path_erode = cv2.erode(path, kernel); #cv2.imshow('Erode by the same amount of pixels',path_erode)#absdiff path = cv2.absdiff(path, path_erode); #cv2.imshow('absdiff',path)#solution channels = cv2.split(img); channels[0] &= ~path; channels[1] &= ~path; channels[2] |= path; dst = cv2.merge(channels); cv2.imshow("solution", dst);#waiting for any key to close windows cv2.waitKey(0) cv2.destroyAllWindows()if __name__ == '__main__': image = sys.argv[-1] SolvingMaze(image)

更多关于Python相关内容可查看本站专题：《Python游戏开发技巧总结》、《Python数据结构与算法教程》、《Python Socket编程技巧总结》、《Python函数使用技巧总结》、《Python字符串操作技巧汇总》、《Python入门与进阶经典教程》及《Python文件与目录操作技巧汇总》

希望本文所述对大家Python程序设计有所帮助。