Ruby实现的最优二叉查找树算法

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我是靠谱客的博主缓慢小熊猫，最近开发中收集的这篇文章主要介绍Ruby实现的最优二叉查找树算法，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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。

复制代码代码如下:

#encoding: utf-8
=begin
author: xu jin
date: Nov 11, 2012
Optimal Binary Search Tree
to find by using EditDistance algorithm
refer to <<introduction to algorithms>>
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