Ruby實現的最優二叉查找樹算法

 　　這篇文章主要介紹了Ruby實現的最優二叉查找樹算法,本文直接給出實現代碼,需要的朋友可以參考下

　　算法導論上的偽碼改寫而成，加上導論的課後練習第一題的解的構造函數。

　　代碼如下:

　　#encoding: utf-8

　　=begin

　　author: xu jin

　　date: Nov 11, 2012

　　Optimal Binary Search Tree

　　to find by using EditDistance algorithm

　　refer to <>

　　example output:

　　"k2 is the root of the tree."

　　"k1 is the left child of k2."

　　"d0 is the left child of k1."

　　"d1 is the right child of k1."

　　"k5 is the right child of k2."

　　"k4 is the left child of k5."

　　"k3 is the left child of k4."

　　"d2 is the left child of k3."

　　"d3 is the right child of k3."

　　"d4 is the right child of k4."

　　"d5 is the right child of k5."

　　The expected cost is 2.75.

　　=end

　　INFINTIY = 1 / 0.0

　　a = ['', 'k1', 'k2', 'k3', 'k4', 'k5']

　　p = [0, 0.15, 0.10, 0.05, 0.10, 0.20]

　　q = [0.05, 0.10, 0.05, 0.05, 0.05 ,0.10]

　　e = Array.new(a.size + 1){Array.new(a.size + 1)}

　　root = Array.new(a.size + 1){Array.new(a.size + 1)}

　　def optimalBST(p, q, n, e, root)

　　w = Array.new(p.size + 1){Array.new(p.size + 1)}

　　for i in (1..n + 1)

　　e[i][i - 1] = q[i - 1]

　　w[i][i - 1] = q[i - 1]

　　end

　　for l in (1..n)

　　for i in (1..n - l + 1)

　　j = i + l -1

　　e[i][j] = 1 / 0.0

　　w[i][j] = w[i][j - 1] + p[j] + q[j]

　　for r in (i..j)

　　t = e[i][r - 1] + e[r + 1][j] + w[i][j]

　　if t < e[i][j]

　　e[i][j] = t

　　root[i][j] = r

　　end

　　end

　　end

　　end

　　end

　　def printBST(root, i ,j, signal)

　　return if i > j

　　if signal == 0

　　p "k#{root[i][j]} is the root of the tree."

　　signal = 1

　　end

　　r = root[i][j]

　　#left child

　　if r - 1< i

　　p "d#{r - 1} is the left child of k#{r}."

　　else

　　p "k#{root[i][r - 1]} is the left child of k#{r}."

　　printBST(root, i, r - 1, 1 )

　　end

　　#right child

　　if r >= j

　　p "d#{r} is the right child of k#{r}."

　　else

　　p "k#{root[r + 1][j]} is the right child of k#{r}."

　　printBST(root, r + 1, j, 1)

　　end

　　end

　　optimalBST(p, q, p.size - 1, e, root)

　　printBST(root, 1, a.size-1, 0)

　　puts "nThe expected cost is #{e[1][a.size-1]}."