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]}."