The optimal binary lookup tree algorithm implemented by Ruby _ruby Special topic

The pseudo code in the introduction of the algorithm is rewritten, coupled with the constructor of the solution of the first question in the introductory lesson practice.

Copy Code code as follows:

#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 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)
    to 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]
&nbs p;     for R in (i.. j)
        t = e[i][r-1] + e[r + 1][j] + w[i][j]
    &nbs p;   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]}."