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 : 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<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。
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稳了!魔兽国服回归的3条重磅消息!官宣时间再确认!
昨天有一位朋友在大神群里分享,自己亚服账号被封号之后居然弹出了国服的封号信息对话框。
这里面让他访问的是一个国服的战网网址,com.cn和后面的zh都非常明白地表明这就是国服战网。
而他在复制这个网址并且进行登录之后,确实是网易的网址,也就是我们熟悉的停服之后国服发布的暴雪游戏产品运营到期开放退款的说明。这是一件比较奇怪的事情,因为以前都没有出现这样的情况,现在突然提示跳转到国服战网的网址,是不是说明了简体中文客户端已经开始进行更新了呢?
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