本文实例为大家分享了python机器人行走步数问题，供大家参考，具体内容如下

创新互联建站网站建设公司是一家服务多年做网站建设策划设计制作的公司,为广大用户提供了做网站、成都网站建设，成都网站设计，一元广告，成都做网站选创新互联建站，贴合企业需求，高性价比，满足客户不同层次的需求一站式服务欢迎致电。

#! /usr/bin/env python3 
# -*- coding: utf-8 -*- 
# fileName ： robot_path.py 
# author : zoujiameng@aliyun.com.cn 
 
# 地上有一个m行和n列的方格。一个机器人从坐标0,0的格子开始移动，每一次只能向左，右，上，下四个方向移动一格，但是不能进入行坐标和列坐标的数位之和大于k的格子。  
# 例如，当k为18时，机器人能够进入方格（35,37），因为3+5+3+7 = 18。但是，它不能进入方格（35,38），因为3+5+3+8 = 19。请问该机器人能够达到多少个格子？ 
class Robot: 
# 共用接口,判断是否超过K 
  def getDigitSum(self, num): 
    sumD = 0 
    while(num>0): 
      sumD+=num%10 
      num/=10 
    return int(sumD) 
 
  def PD_K(self, rows, cols, K): 
    sumK = self.getDigitSum(rows) + self.getDigitSum(cols) 
    if sumK > K: 
      return False 
    else: 
      return True 
 
  def PD_K1(self, i, j, k): 
    "确定该位置是否可以走,将复杂约束条件设定" 
    index = map(str,[i,j]) 
    sum_ij = 0 
    for x in index: 
      for y in x: 
        sum_ij += int(y) 
    if sum_ij <= k: 
      return True 
    else: 
      return False 
 
# 共用接口,打印遍历的visited二维list 
  def printMatrix(self, matrix, r, c): 
    print("cur location(", r, ",", c, ")") 
    for x in matrix: 
      for y in x:  
        print(y, end=' ') 
      print() 
 
 #回溯法 
  def hasPath(self, threshold, rows, cols): 
    visited = [ [0 for j in range(cols)] for i in range(rows) ] 
    count = 0 
    startx = 0 
    starty = 0 
    #print(threshold, rows, cols, visited) 
    visited = self.findPath(threshold, rows, cols, visited, startx, starty, -1, -1) 
    for x in visited: 
      for y in x: 
        if( y == 1): 
          count+=1 
    print(visited) 
    return count 
 
  def findPath(self, threshold, rows, cols, visited, curx, cury, prex, prey): 
    if 0 <= curx < rows and 0 <= cury < cols and self.PD_K1(curx, cury, threshold) and visited[curx][cury] != 1: # 判断当前点是否满足条件 
      visited[curx][cury] = 1 
    self.printMatrix(visited, curx, cury) 
    prex = curx 
    prey = cury 
    if cury+1 < cols and self.PD_K1(curx, cury+1, threshold) and visited[curx][cury+1] != 1: # east 
      visited[curx][cury+1] = 1 
      return self.findPath(threshold, rows, cols, visited, curx, cury+1, prex, prey) 
    elif cury-1 >= 0 and self.PD_K1(curx, cury-1, threshold) and visited[curx][cury-1] != 1: # west 
      visited[curx][cury-1] = 1 
      return self.findPath(threshold, rows, cols, visited, curx, cury-1, prex, prey) 
    elif curx+1 < rows and self.PD_K1(curx+1, cury, threshold) and visited[curx+1][cury] != 1: # sourth 
      visited[curx+1][cury] = 1 
      return self.findPath(threshold, rows, cols, visited, curx+1, cury, prex, prey) 
    elif 0 <= curx-1 and self.PD_K1(curx-1, cury, threshold) and visited[curx-1][cury] != 1: # north 
      visited[curx-1][cury] = 1 
      return self.findPath(threshold, rows, cols, visited, curx-1, cury, prex, prey) 
    else: # 返回上一层,此处有问题 
      return visited#self.findPath(threshold, rows, cols, visited, curx, cury, prex, prey) 
 #回溯法2 
  def movingCount(self, threshold, rows, cols): 
    visited = [ [0 for j in range(cols)] for i in range(rows) ] 
    print(visited) 
    count = self.movingCountCore(threshold, rows, cols, 0, 0, visited); 
    print(visited) 
    return count 
 
  def movingCountCore(self, threshold, rows, cols, row, col, visited): 
    cc = 0 
    if(self.check(threshold, rows, cols, row, col, visited)):  
      visited[row][col] = 1 
      cc = 1 + self.movingCountCore(threshold, rows, cols, row+1, col,visited) + self.movingCountCore(threshold, rows, cols, row, col+1, visited) + self.movingCountCore(threshold, rows, cols, row-1, col, visited) + self.movingCountCore(threshold, rows, cols, row, col-1, visited) 
    return cc 
 
  def check(self, threshold, rows, cols, row, col, visited): 
    if( 0 <= row < rows and 0 <= col < cols and (self.getDigitSum(row)+self.getDigitSum(col)) <= threshold and visited[row][col] != 1):  
      return True; 
    return False  
 
# 暴力法，直接用当前坐标和K比较 
  def force(self, rows, cols, k): 
    count = 0 
    for i in range(rows): 
      for j in range(cols): 
        if self.PD_K(i, j, k): 
          count+=1 
    return count 
# 暴力法2, 用递归法来做 
  def block(self, r, c, k):  
    s = sum(map(int, str(r)+str(c))) 
    return s>k 
  def con_visited(self, rows, cols): 
    visited = [ [0 for j in range(cols)] for i in range(rows) ] 
    return visited 
  def traval(self, r, c, rows, cols, k, visited): 
    if not (0<=r
另外有需要云服务器可以了解下创新互联scvps.cn，海内外云服务器15元起步，三天无理由+7*72小时售后在线，公司持有idc许可证，提供“云服务器、裸金属服务器、高防服务器、香港服务器、美国服务器、虚拟主机、免备案服务器”等云主机租用服务以及企业上云的综合解决方案，具有“安全稳定、简单易用、服务可用性高、性价比高”等特点与优势，专为企业上云打造定制，能够满足用户丰富、多元化的应用场景需求。
                

                网页题目：python机器人行走步数问题的解决-创新互联                

                路径分享：http://ybzwz.com/article/idpgp.html