Question

我正在编写一个基于BFS的代码，其中必须计算与标记为1的框的最近距离。标记为1的框的距离为0。

这是Google kickstart 2019回合A：包裹中的问题。

链接：https://codingcompetitions.withgoogle.com/kickstart/round/0000000000050e01/000000000006987d

import time 
def multibfs (office, r, c):

    #office is a 2-D list with 0's and 1's
    visited = [[False]*c for x in range (r)] 
    values = [[1e9]*c for x in range (r)] #closest distance

    queue = [] #Bfs queue

    for i in range (r): 
        for j in range (c): 
            if office[i][j]==1: 
                visited[i][j]=True 
                values[i][j]=0 
                queue.append([i,j]) 

    while queue: 
        s = queue.pop(0)
        iPos = s[0]
        jPos = s[1]

        for i in range (iPos-1, iPos+2):
            if i in range (0, r): 
                if visited[i][jPos] == False:
                    queue.append([i, jPos])
                    visited[i][jPos] = True
                    values[i][jPos] = values[iPos][jPos] + 1

        for j in range (jPos-1, jPos+2):
            if j in range (0, c):
                if visited[iPos][j] == False:
                    queue.append([iPos, j])
                    visited[iPos][j] = True
                    values[iPos][j] = values[iPos][jPos] + 1

    return values 

tic=time.time() 
o = [[1]*250 for x in range (250)]
val = multibfs(o, 250, 250) 
print(time.time()-tic)

对于r和c等于250，并且office只有1的情况，python3.6的执行时间大约为0.7s。我希望PyPy的实现会更快，但是会花费更长的时间

为什么PyPy实现比python运行慢得多？

0 个答案: