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(",",c,")") for x in matrix: for y in x: print(y,end=' ') print() #回溯法 def hasPath(self,threshold,cols): visited = [ [0 for j in range(cols)] for i in range(rows) ] count = 0 startx = 0 starty = 0 #print(threshold,visited) visited = self.findpath(threshold,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,curx,cury,prex,prey): if 0 <= curx < rows and 0 <= cury < cols and self.PD_K1(curx,threshold) and visited[curx][cury] != 1: # 判断当前点是否满足条件 visited[curx][cury] = 1 self.printMatrix(visited,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,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,prey) elif curx+1 < rows and self.PD_K1(curx+1,threshold) and visited[curx+1][cury] != 1: # sourth visited[curx+1][cury] = 1 return self.findpath(threshold,curx+1,prey) elif 0 <= curx-1 and self.PD_K1(curx-1,threshold) and visited[curx-1][cury] != 1: # north visited[curx-1][cury] = 1 return self.findpath(threshold,curx-1,prey) else: # 返回上一层,此处有问题 return visited#self.findpath(threshold,prey) #回溯法2 def movingCount(self,cols): visited = [ [0 for j in range(cols)] for i in range(rows) ] print(visited) count = self.movingCountCore(threshold,visited); print(visited) return count def movingCountCore(self,row,col,visited): cc = 0 if(self.check(threshold,visited)): visited[row][col] = 1 cc = 1 + self.movingCountCore(threshold,row+1,visited) + self.movingCountCore(threshold,col+1,row-1,col-1,visited) return cc def check(self,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,k): count = 0 for i in range(rows): for j in range(cols): if self.PD_K(i,k): count+=1 return count # 暴力法2,用递归法来做 def block(self,k): s = sum(map(int,str(r)+str(c))) return s>k def con_visited(self,cols): visited = [ [0 for j in range(cols)] for i in range(rows) ] return visited def traval(self,k,visited): if not (0<=r<rows and 0<=c<cols): return if visited[r][c] != 0 or self.block(r,k): visited[r][c] = -1 return visited[r][c] = 1 global acc acc+=1 self.traval(r+1,visited) self.traval(r,c+1,visited) self.traval(r-1,c-1,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,o.con_visited(m,n))) print(o.movingCount(k,n))
