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 =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:[5,8,2,9] The optimum order is:((A1(A2A3))(A4A5)) The total number of multiplications is: 388 =end INFINTIY = 1 / 0.0 p = [30,25] 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,s[i][j]) print_optimal_parens(s,s[i][j] + 1,j) print ")" end end def process(p,s) matrix_chain_order(p,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,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,Array.new(p.size){Array.new(p.size)} process(p,s)
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