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#! /usr/bin/env python3 # -*- coding: utf-8 -*- # fileName : robot_path.py # author : zoujiameng@aliyun.com.cn # 地上有一個m行和n列的方格。一個機器人從坐標(biāo)0,0的格子開始移動,每一次只能向左,右,上,下四個方向移動一格,但是不能進(jìn)入行坐標(biāo)和列坐標(biāo)的數(shù)位之和大于k的格子。 # 例如,當(dāng)k為18時,機器人能夠進(jìn)入方格(35,37),因為3+5+3+7 = 18。但是,它不能進(jìn)入方格(35,38),因為3+5+3+8 = 19。請問該機器人能夠達(dá)到多少個格子? class Robot: # 共用接口,判斷是否超過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): "確定該位置是否可以走,將復(fù)雜約束條件設(shè)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,打印遍歷的visited二維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: # 判斷當(dāng)前點是否滿足條件 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: # 返回上一層,此處有問題 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,直接用當(dāng)前坐標(biāo)和K比較 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, 用遞歸法來做 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關(guān)于“python如何實現(xiàn)機器人行走效果”這篇文章就分享到這里了,希望以上內(nèi)容可以對大家有一定的幫助,使各位可以學(xué)到更多知識,如果覺得文章不錯,請把它分享出去讓更多的人看到。
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本文標(biāo)題:python如何實現(xiàn)機器人行走效果-創(chuàng)新互聯(lián)
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