這篇文章主要介紹了Ruby實現的矩陣連乘算法,本文直接給出實現代碼,需要的朋友可以參考下
動態規劃解決矩陣連乘問題，隨機產生矩陣序列，輸出形如((A1(A2A3))(A4A5))的結果。
代碼：
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 #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: %dnn", 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)
