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./
2./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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希望本文所述对大家Python程序设计有所帮助。
时间： -09-23
