두 개 이상의 시퀀스를 부분 시퀀스로 포함하는 가장 짧은 슈퍼 시퀀스를 찾는 C++ 프로그램

여기에서는 두 개 이상의 시퀀스를 부분 시퀀스로 포함하는 최단 시퀀스를 찾는 C++ 프로그램에 대해 설명합니다.

알고리즘

Begin
   function ShortestSubSeq() returns supersequence of A and B:
   1) Declare an array ss[i][j] which contains length of shortest supersequence for A[0 .. i-1] and B[0 .. j-1].
   2) Find the length of the possible supersequence in bottom up manner using recursion.
   3) Declare an array ss[i][j] which stores length of shortest supersequence for A[0 .. i-1] and B[0 .. j-1] in index.
   4) Declare a string s to store the shortest subsequence.
   5) Initialize i = a, j = b.
   6) while (i > 0 and j > 0)
      A) If current character in A and B are same, then current character is part of shortest supersequence.
      Put current character in result.
      Reduce values of i, j and index.
      B) Else if
      If current character in A and B are different,
      Put current character of B in result.
      Reduce values of j and index.
      C) Else
      Put current character of A in result.
      Reduce values of i and index.
   7) While (i > 0)
   Put remaining characters of A in the result string.
   8) While(j>0)
   Put remaining characters of B in the result string.
   9) Reverse the string and return it.
End

예시

#include <bits/stdc++.h>
using namespace std;
string ShortestSuperSeq(string A, string B) {
   int a = A.length();  
   int b = B.length();
   int ss[a + 1][b + 1];
   for (int i = 0; i <= a; i++) {
      for (int j = 0; j <= b; j++) {
         if(i == 0)
            ss[i][j] = j;
         else if(j == 0)
            ss[i][j] = i;
         else if(A[i - 1] == B[j - 1])
            ss[i][j] = 1 + ss[i - 1][j - 1];
         else
            ss[i][j] = 1 + min(ss[i - 1][j], ss[i][j - 1]);
      }
   }
   int index = ss[a][b];
   string s;
   int i = a, j = b;
   while (i > 0 && j > 0) {
      //If current character in A and B are same, then current character is part of shortest supersequence
      if (A[i - 1] == B[j - 1]) {
         //Put current character in result.
         //reduce values of i, j and index.
         s.push_back(A[i - 1]);
         i--, j--, index--;
      }
      //If current character in A and B are different,
      else if (ss[i - 1][j] > ss[i][j - 1]) {
         //Put current character of B in result.
         //reduce values of j and index.
         s.push_back(B[j - 1]);
         j--, index--;
      }
      //Put current character of A in result.
      //reduce values of i and index.
      else {
         s.push_back(A[i - 1]);
         i--, index--;
      }
   }
   //put remaining characters of A in the result string.
   while (i > 0) {
      s.push_back(A[i - 1]);
      i--, index--;
   }
   //put remaining characters of B in the result string.
   while (j > 0) {
      s.push_back(B[j - 1]);
      j--, index--;
   }
   reverse(s.begin(), s.end()); //Reverse the string and return it.
   return s;
}
int main() {
   string M = "ABBCDDEEFF";
   string N = "ABCDEEEFF";
   cout <<"The Shortest SuperSequence is:"<< ShortestSuperSeq(M, N);
   return 0;
}

출력

The Shortest SuperSequence is:ABBCDEDEEFF