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)







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