Ruby如何实现最优二叉查找树算法

来源:爱站网时间:2020-10-22编辑:网友分享
二叉查找树算法在Ruby中是非常常用的算法,那么你知道Ruby如何实现最优二叉查找树算法吗?感兴趣的小伙伴们就跟爱站小编一起去看看吧。

二叉查找树算法在Ruby中是非常常用的算法,那么你知道Ruby如何实现最优二叉查找树算法吗?感兴趣的小伙伴们就跟爱站小编一起去看看吧。

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 >
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] = 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     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]}."

 

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