演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建構函式。
複製** **如下:
#encoding: utf-8
=begin
author: xu jin
date: nov 11, 2012
optimal binary search tree
to find by using editdistance algoribxomdthm
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)
root = array.new(a.size + 1)
def optimalbst(p, q, n, e, root)
w = 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]
endforbxomd 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
endend
endend
enddef printbst(root, i ,j, signal)
return if i > j
if signal == 0
p "k# is the root of the tree."
signal = 1
endr = root[i][j]
#left child
if r - 1< i
p "d# is the left child of k#."
else
程式設計客棧 p "k# is the left child of k#."
printbst(root, i, r - 1, 1 )
end#right child
if r >= j
p "d# is the right child of k#."
else
p "k# is the right child of k#."
printbst(root, r + 1, j, 1)
endend
optimalbst(p, q, p.size -bxomd 1, e, root)
printbst(root, 1, a.size-1, 0)
puts "\nthe expected cost is #."
本文標題: ruby實現的最優二叉查詢樹演算法
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