算法導(dǎo)論上的偽碼改寫而成，加上導(dǎo)論的課后練習(xí)第一題的解的構(gòu)造函數(shù)。
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]}."
標(biāo)簽：三沙 遼寧 荊門 普洱 張家界 公主嶺 永州 梧州
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