Python基于分水岭算法解决走迷宫游戏示例

Elizabeth ·

更新时间:2024-11-10

· 598 次阅读

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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分水岭算法迷宫迷宫游戏示例算法 Python

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