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]}."
