Python基于分水岭算法解决走迷宫游戏示例

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 transformation
needed module:cv2/numpy/sys
ref:
1.http://www.mazegenerator.net/
2.http://blog.leanote.com/post/leeyoung/539a629aab35bc44e2000000
@author:Robin Chen
"""
import cv2
import numpy as np
import sys
def 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)

希望本文所述对大家Python程序设计有所帮助。

来源：http://blog.csdn.net/chendejiao/article/details/43231659