python -- 机器人行走步数问题

#! /usr/bin/env python3# -*- coding: utf-8 -*-# fileName ： robot_path.py# author : [e-mail]# 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Kdef getDigitSum(self, num):sumD = 0while(num>0):sumD+=num%10num/=10return int(sumD)def PD_K(self, rows, cols, K):sumK = self.getDigitSum(rows) + self.getDigitSum(cols)if sumK > K:return Falseelse:return Truedef 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 = 0for x in index:for y in x:sum_ij += int(y)if sum_ij <= k:return Trueelse:return False# 共用接口,打印遍历的visited二维listdef 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 = 0startx = 0starty = 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+=1print(visited)return countdef 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] = 1self.printMatrix(visited, curx, cury)prex = curxprey = curyif cury+1 < cols and self.PD_K1(curx, cury+1, threshold) and visited[curx][cury+1] != 1: # eastvisited[curx][cury+1] = 1return 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: # westvisited[curx][cury-1] = 1return 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: # sourthvisited[curx+1][cury] = 1return 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: # northvisited[curx-1][cury] = 1return 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2def 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 countdef movingCountCore(self, threshold, rows, cols, row, col, visited):cc = 0if(self.check(threshold, rows, cols, row, col, visited)):  visited[row][col] = 1cc = 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 ccdef 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 = 0for i in range(rows):for j in range(cols):if self.PD_K(i, j, k):count+=1return 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>kdef con_visited(self, rows, cols):visited = [ [0 for j in range(cols)] for i in range(rows) ]return visiteddef traval(self, r, c, rows, cols, k, visited):if not (0<=r<rows and 0<=c<cols):returnif visited[r][c] != 0 or self.block(r, c, k):visited[r][c] = -1returnvisited[r][c] = 1global accacc+=1self.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 accif __name__ == "__main__":# 调用测试m = 3n = 3k = 1o = Robot()print(o.hasPath(k, m, n))print(o.force(m,n,k))global accacc = 0print(o.traval(0, 0, m, n, k, o.con_visited(m,n)))print(o.movingCount(k, m, n))