Longest Common subsequence of LCS and LCS common subsequence

Longest Common subsequence of LCS and LCS common subsequence

Problem: The Longest Common subsequence does not require that the obtained string be continuous in the given string. For example, the input two strings ABCBDAB and BDCABA, And the strings BCBA and BDAB are both their public longest headers.

This is a dynamic planning problem.

Answer: set sequence X = <x0, x1 ,..., xm> and Y = <y0, y1 ,..., one of the longest common subsequences of yn> is Z = <z0, z1 ,..., zk>, then:

1) if xm = yn, there must be zk = xm = yn, then zk-1 is the longest common subsequence of xm-1 and yn-1;

2) If xm =yn and zk =xm, Z is the longest common subsequence of Xm-1 and Y;

3) If xm =yn and zk =yn, Z is the longest common subsequence of X and Yn-1;

That is to say:

When xm = yn, LCS (Xm, Yn) = LCS (Xm-1, Yn-1) + 1;

When xm =yn, LCS (Xm, yn) = max {LCS (Xm-1, Yn), LCS (Xm, Yn-1 )};

When X and Y are empty, the LCS length is 0;

 

1 # include <iostream> 2 # include <cmath> 3 # define INF 9999999 4 using namespace std; 5 int a [100] [100]; // record the computed subproblem 6 int fun (const char * str1, const char * str2, int I, int j) 7 {8 if (a [I] [j] <INF) return a [I] [j]; // 9 else if (I = 0 | j = 0) a [I] [j] = 0; 10 else if (str1 [I-1] = str2 [J-1]) a [I] [j] = fun (str1, str2, I-1, J-1) + 1; 11 else a [I] [j] = max (fun (str1, str2, I-1, j), fun (str1, str2, I, J-1 )); 12 return a [I] [j]; 13} 14 int longest (const char * str1, const char * str2) 15 {16 int m = strlen (str1 ); 17 int n = strlen (str2); 18 memset (a, INF, sizeof (a); // set element a to INF19 return fun (str1, str2, m, n); 20} 21 int main () 22 {23 char * str1 = new char [100], * str2 = new char [100]; 24 cout <"input first string:"; 25 cin> str1; 26 cout <"input second string:"; 27 cin> str2; 28 cout <"the longest length of sunstring is:"; 29 cout <longest (str1, str2) <endl; 30 delete [] str1; 31 delete [] str2; 32}

 

Result: