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我想使用BFS实现搜索。算法说我必须使用队列才能获得 FIFO 效果。我阅读了 Chris Okasaki 的Purely Functional Data Structures书，发现了如何创建队列（我使用 F# 编写的）：

type 'a queue = 'a list * 'a list
let emtpy = [],[]
let isEmpty = function
    | [],_ -> true
    | _ -> false

let checkf = function
    | [],r -> List.rev r,[]
    | q -> q

let snoc (f,r) x = checkf (f,x :: r)

let head = function
    | ([],_) -> failwith "EMPTY"
    | (x::f,r) -> x

let tail = function
    | ([],_) -> failwith "EMPTY"
    | (x::f,r) -> checkf (f,r)

任何人都知道如何将其实施到 BFS？

我有这段代码可以从列表中创建一棵树：

let data = [4;3;8;7;10;1;9;6;5;0;2]

type Tree<'a> = 
    | Node of Tree<'a> * 'a * Tree<'a>
    | Leaf

let rec insert tree element = 
    match element,tree with
    | x,Leaf        -> Node(Leaf,x,Leaf)
    | x,Node(l,y,r) when x <= y -> Node((insert l x),y,r)
    | x,Node(l,y,r) when x > y -> Node(l,y,(insert r x))
    | _ -> Leaf

let makeTree = List.fold insert Leaf data

（想结合这两个代码）

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1 回答 1

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BFS算法是这样的：

Initialise the search by placing the starting vertex in the queue.
While the queue is not empty.
  Remove the front vertex from the queue.
  If this is a solution then we're finished -- report success.
  Otherwise, compute the immediate children of this vertex and enqueue them.
Otherwise we have exhausted the queue and found no solution -- report failure.

我的 F# 语法有点摇摆不定，但这是我草拟解决方案的方式：

bfs start = bfsLoop ([start], [])

bfsLoop q0 =
  if   isEmpty q0
  then failWith "No solution"
  else v = head q0
       if   isSolution v
       then v
       else q1 = tail q0
            vs = childrenOf v
            q = foldl snoc vs q1
            bfsLoop q

希望这可以帮助。

于 2011-05-20T06:31:20.687 回答