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：

#encoding: utf-8=beginauthor: xu jin, 4100213date: Oct 28, 2012MatrixChainto find an optimum order by using MatrixChain algorithmexample 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: 15125The random array is：[5, 8, 8, 2, 5, 9]The optimum order is:((A1(A2A3))(A4A5))The total number of multiplications is: 388 =endINFINTIY = 1 / 0.0p = [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 endend 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 ")" endenddef 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])endputs "The given array is:" + p.to_sprocess(p, m, s)#produce a random arrayp = Array.newx = rand(10)(0..x).each{|index| p[index] = rand(10) + 1}puts "The random array is：" + p.to_sm, s = Array.new(p.size){Array.new(p.size)}, Array.new(p.size){Array.new(p.size)}process(p, m, s)