Hi，大家好，我是编程小6，很荣幸遇见你，我把这些年在开发过程中遇到的问题或想法写出来，今天说一说python数字推盘游戏怎么显示步数_python机器人行走步数问题的解决,希望能够帮助你!!!。

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

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我们。

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时间： 2018-01-26

今天的分享到此就结束了，感谢您的阅读，如果确实帮到您，您可以动动手指转发给其他人。