星鳕鱼问题

Question

我有一个问题，使我的明星实施工作。你能指路吗？谢谢您。在

#!/usr/bin/env python

import sys, re, math  

grid = []  
g = []
h = []

width =  int(sys.argv[1])
height = int(sys.argv[2])

open = []
close = []

startpos = (0,0) #(height,width)
endpos = (6,5) #(height,width)

#Functions

def findlowestcostinopen():
 lowest = 9999
 lowestpair = []
 for q in open:
  sum = int(g[q[0]][q[1]])+int(h[q[0]][q[1]])
  #print sum,lowest
  if sum<lowest:
   lowestpair = q
   lowest=sum

 return lowestpair
# Init

for q in range(height):
 temp = []
 for w in range(width):
  temp.append((0,0))
 grid.append(temp)

for q in range(height):
 temp = []
 for w in range(width):
  temp.append(0)
 g.append(temp)

for q in range(height):  
 temp = []  
 for w in range(width):    
  temp.append(0)  
 h.append(temp)  



for q in range(height):  
 for w in range(width):  
  h[q][w]=abs(endpos[0]-q)*10 + abs(endpos[1]-w)*10  

open.append(startpos)  
switch = True  
while switch and open:  
 #Find the smallest cost  
 lowestcost = findlowestcostinopen()  
 print lowestcost,endpos  
 if lowestcost == endpos:  
  switch = False  
  print 'found',lowestcost  


 parentgcost=int(g[lowestcost[0]][lowestcost[1]])    
 #print parentgcost  
 #Check every directly connected node     
 for q in range(-1,2):    
  for w in range(-1,2):    
   currentnode = ((lowestcost[0]+q),(lowestcost[1]+w))  
   if q==0 and w==0:  
    ''''''
   elif(currentnode[0]<0 or currentnode[0]>(height-1)):  
    '''Vertical out'''  
   elif(currentnode[1]<0 or currentnode[1]>(width-1)):  
    '''Horizontal out'''  
   elif(grid[currentnode[0]][currentnode[1]]=='wall'):
    '''WALL'''  
   elif open.count((currentnode[0],currentnode[1]))>0: 

    ''''''  
    currentg = g[currentnode[0]][currentnode[1]]    

    if (q==0 and w==1) or (q==0 and w==-1) or (q==1 and w==0) or (q==-1 and w==0):  
     newsum = parentgcost+10  
    else: newsum = parentgcost+14  

    if newsum<currentg:  
     g[currentnode[0]][currentnode[1]]=newsum  

    grid[currentnode[0]][currentnode[1]]=lowestcost   

   elif close.count((currentnode[0],currentnode[1]))>0:  
    '''EXISTS IN CLOSE'''  
   else:   
    #Time to calculate g values  
    if q==0:
     if w==-1 or w==1:
      nodecost = parentgcost+10
    elif q==1:  
     if w==0:  
      nodecost = parentgcost+10  
     else:   
      nodecost = parentgcost+14  
    elif q==-1:  
     if w==0:  
      nodecost = parentgcost+10  
     else:  
      nodecost = parentgcost+14  
    g[(currentnode[0])][(currentnode[1])]=nodecost   
    grid[(currentnode[0])][(currentnode[1])]=lowestcost  
    #print nodecost  

    open.append(currentnode)  

Answer 1

这是一个*搜索的伪代码。它很容易被转录成Python。在

Wikipedia A* search page

Answer 2

一些问题：

1. 您这样做是为了征求意见：

   '''some text'''
   

   这实际上不是一个注释，而是一个字符串。你只是不把它分配给任何东西。请改为发表评论：

   # some text
   

2. 这段代码很难阅读：

           if q==0:
            if w==-1 or w==1:
             nodecost = parentgcost+10
           elif q==1:  
            if w==0:  
             nodecost = parentgcost+10  
            else:   
             nodecost = parentgcost+14  
           elif q==-1:  
            if w==0:  
             nodecost = parentgcost+10  
            else:  
             nodecost = parentgcost+14  
   

   更改为：

           if q==0 and (w==-1 or w==1):
               nodecost = parentgcost+10
           elif q==1 and w==0:  
               nodecost = parentgcost+10  
           elif q==1:  
               nodecost = parentgcost+14  
           elif q==-1 and w==0:  
               nodecost = parentgcost+10  
           elif q==-1:  
               nodecost = parentgcost+14  
   

   请注意如何使用四个空格来缩进，而不仅仅是一个。

3. 此处不需要括号：

           g[(currentnode[0])][(currentnode[1])]=nodecost   
   

   更改为

           g[currentnode[0]][currentnode[1]]=nodecost   
   

4. 你太喜欢索引了。这也让人难以阅读。在

           g[(currentnode[0])][(currentnode[1])]=nodecost
   

   最好是

           height, width = currentnode
           g[height][width] = nodecost
   

这些都不能解决您的问题，因为您没有说明这是什么，甚至代码应该做什么。在