如何在大空间尺度上加速A *算法？

Question

如何在大空间尺度上加速A *算法？

从http://ccl.northwestern.edu/netlogo/models/community/Astardemo，我通过使用网络中的节点来定义最低成本路径来编码A *算法。代码似乎有效，但是当我在大空间尺度上使用它时它太慢了。我的景观有1000个补丁x 1000个补丁，1个补丁= 1个像素。即使我减少400补丁x 400补丁1补丁= 1像素，它仍然太慢（我不能修改我的景观低于400补丁x 400补丁）。这是代码：

to find-path [ source-node destination-node] 

let search-done? false

let search-path []

let current-node 0

set list-open []

set list-closed []  

let list-links-with-nodes-in-list-closed []

let list-links []

set list-open lput source-node list-open

while [ search-done? != true]

[    

ifelse length list-open != 0

[

  set list-open sort-by [[f] of ?1 < [f] of ?2] list-open 

  set current-node item 0 list-open 

  set list-open remove-item 0 list-open 

  set list-closed lput current-node list-closed

  ask current-node

  [  

    if parent-node != 0[

    set list-links-with-nodes-in-list-closed lput link-with parent-node list-links-with-nodes-in-list-closed 

    ]

    ifelse any? (nodes-on neighbors4) with [ (xcor = [ xcor ] of destination-node) and (ycor = [ycor] of destination-node)]

    [

      set search-done? true 

    ]

    [        

      ask (nodes-on neighbors4) with [ (not member? self list-closed) and (self != parent-node) ]  

      [  

        if not member? self list-open and self != source-node and self != destination-node

        [

          set list-open lput self list-open

          set parent-node current-node

          set list-links sentence (list-links-with-nodes-in-list-closed) (link-with parent-node)

          set g sum (map [ [link-cost] of ? ] list-links)

          set h distance destination-node 

          set f (g + h)

        ]

      ]

    ]

  ]

]

[

  user-message( "A path from the source to the destination does not exist." )

  report []

 ]

]

set search-path lput current-node search-path

let temp first search-path

while [ temp != source-node ]

[

 ask temp

[

  set color red

]

不幸的是，我不知道如何加快这段代码。是否有解决方案在大空间尺度上快速计算最低成本路径？