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
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]}."
Ruby,二叉查找树,算法
稳了!魔兽国服回归的3条重磅消息!官宣时间再确认!
昨天有一位朋友在大神群里分享,自己亚服账号被封号之后居然弹出了国服的封号信息对话框。
这里面让他访问的是一个国服的战网网址,com.cn和后面的zh都非常明白地表明这就是国服战网。
而他在复制这个网址并且进行登录之后,确实是网易的网址,也就是我们熟悉的停服之后国服发布的暴雪游戏产品运营到期开放退款的说明。这是一件比较奇怪的事情,因为以前都没有出现这样的情况,现在突然提示跳转到国服战网的网址,是不是说明了简体中文客户端已经开始进行更新了呢?
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