python机器人行走步数问题的解决

Python  /  管理员 发布于 7年前   246

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

#! /usr/bin/env python3 # -*- coding: utf-8 -*- # fileName ： robot_path.py # author : [email protected]  # 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<rows and 0<=c<cols):       return     if visited[r][c] != 0 or self.block(r, c, k):       visited[r][c] = -1       return     visited[r][c] = 1     global acc     acc+=1     self.traval(r+1, c, rows, cols, k, visited)     self.traval(r, c+1, rows, cols, k, visited)     self.traval(r-1, c, rows, cols, k, visited)     self.traval(r, c-1, rows, cols, k, visited)     return acc  if __name__ == "__main__":   # 调用测试   m = 3   n = 3   k = 1   o = Robot()   print(o.hasPath(k, m, n))   print(o.force(m,n,k))   global acc   acc = 0   print(o.traval(0, 0, m, n, k, o.con_visited(m,n)))   print(o.movingCount(k, m, n)) 

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持。