最長公用子序列的Ruby實現

最長公用子序列，LCS，動態規劃實現。

#encoding: utf-8#author: xu jin, 4100213#date: Nov 01, 2012#Longest-Commom-Subsequence#to find a longest commom subsequence of two given character arrays by using LCS algorithm#example output:#The random character arrays are: ["b", "a", "c", "a", "a", "b", "d"] and ["a", "c", "a", "c", "a", "a", "b"]#The Longest-Commom-Subsequence is: a c a a bchars = ("a".."e").to_ax, y = [], []1.upto(rand(5) + 5) { |i| x << chars[rand(chars.size-1)] }1.upto(rand(5) + 5) { |i| y << chars[rand(chars.size-1)] }printf("The random character arrays are: %s and %s\n", x, y)c = Array.new(x.size + 1){Array.new(y.size + 1)}b = Array.new(x.size + 1){Array.new(y.size + 1)}def LCS_length(x, y ,c ,b)       m, n = x.size, y.size      (0..m).each{|i| c[i][0] = 0}      (0..n).each{|j| c[0][j] = 0}      for i in (1..m) do       for j in(1..n) do        if(x[i - 1] == y [j - 1])          c[i][j] = c[i - 1][j - 1] + 1;          b[i][j] = 0        else          if(c[i - 1][j] >= c[i][j - 1])            c[i][j] = c[i - 1][j]            b[i][j] = 1          else            c[i][j] = c[i][j - 1]            b[i][j] = 2          end        end      end     endenddef Print_LCS(x, b, i, j)    return if(i == 0 || j == 0)    if(b[i][j] == 0)        Print_LCS(x, b, i-1, j-1)        printf("%c ", x[i - 1])    elsif(b[i][j] == 1)        Print_LCS(x, b, i-1, j)    else        Print_LCS(x, b, i, j-1)    endendLCS_length(x, y, c ,b) print "The Longest-Commom-Subsequence is: "Print_LCS(x, b, x.size, y.size)