Question

我想使用BFS实现搜索。算法说我必须使用队列来获得FIFO效果。我读了Chris Okasaki的Purely Functional Data Structures书并找到了如何排队（我用F＃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？

我有这个代码从列表中生成一个树：

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

（想要结合这两个代码）

Answer 1

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＃语法有点不稳定，但这是我如何勾画出解决方案：

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

希望这有帮助。

F＃中的广度优先搜索（BFS）

1 个答案: