动态规划解决矩阵连乘问题，随机产生矩阵序列，输出形如((A1(A2A3))(A4A5))的结果。
代码：
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#encoding: utf-8
=begin
author: xu jin, 4100213
date: Oct 28, 2012
MatrixChain
to find an optimum order by using MatrixChain algorithm
example output:
The given array is:[30, 35, 15, 5, 10, 20, 25]
The optimum order is:((A1(A2A3))((A4A5)A6))
The total number of multiplications is: 15125
The random array is：[5, 8, 8, 2, 5, 9]
The optimum order is:((A1(A2A3))(A4A5))
The total number of multiplications is: 388
=end
INFINTIY = 1 / 0.0
p = [30, 35, 15, 5, 10, 20, 25]
m, s = Array.new(p.size){Array.new(p.size)}, Array.new(p.size){Array.new(p.size)}
def matrix_chain_order(p, m, s)
n = p.size - 1
(1..n).each{|i| m[i][i] = 0}
for r in (2..n) do
for i in (1..n - r + 1) do
j = r + i - 1
m[i][j] = INFINTIY
for k in (i...j) do
q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j]
m[i][j], s[i][j] = q, k if(q < m[i][j])
end
end
end
end
def print_optimal_parens(s, i, j)
if(i == j) then
print "A" + i.to_s
else
print "("
print_optimal_parens(s, i, s[i][j])
print_optimal_parens(s, s[i][j] + 1, j)
print ")"
end
end
def process(p, m, s)
matrix_chain_order(p, m, s)
print "The optimum order is:"
print_optimal_parens(s, 1, p.size - 1)
printf("\nThe total number of multiplications is: %d\n\n", m[1][p.size - 1])
end
puts "The given array is:" + p.to_s
process(p, m, s)
#produce a random array
p = Array.new
x = rand(10)
(0..x).each{|index| p[index] = rand(10) + 1}
puts "The random array is：" + p.to_s
m, s = Array.new(p.size){Array.new(p.size)}, Array.new(p.size){Array.new(p.size)}
process(p, m, s)
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