这篇文章主要介绍了Ruby实现的最优二叉查找树算法,本文直接给出实现代码,需要的朋友可以参考下

　　算法导论上的伪码改写而成，加上导论的课后练习第一题的解的构造函数。

　　代码如下:

　　#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]}."