最優二叉尋找樹的Ruby實現

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。

#encoding: utf-8=beginauthor: xu jindate: Nov 11, 2012Optimal Binary Search Treeto find by using EditDistance algorithmrefer 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.  =endINFINTIY = 1 / 0.0a = ['', '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  endenddef 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  endoptimalBST(p, q, p.size - 1, e, root)printBST(root, 1, a.size-1, 0)puts "\nThe expected cost is #{e[1][a.size-1]}."