Longest Common subsequence Problem

The longest public subsequence problem has been seen in many forums for a long time. I saw a post posted a few days ago and I went to the dynamic planning section in the introduction to algorithms. I will not elaborate on this issue, paste the specific code of the C ++ implementation directly. [Cpp] // large public subsequence # pragma once # include <string> using std: string; # define OVER 1 // indicated by an arrow in the book, here, use macros instead of # define LEFT 2 # define LEFTOVER 3 class LCS {private: string sX; // used to store the sub-sequence string sY; int ** c; // these two-dimensional pointers are used to maintain a dynamic table. The longest common subsequence int ** B; int m will be generated recursively Based on the table; // represents the length of the two strings respectively. In order to be consistent with the pseudocode in the book, the int n; public: LCS ();~ LCS (); void ProRun (); // only the dual-loop void Running () of the dynamic planning part; // for main () to call void PrintResult (int, int ); // recursive output}; [cpp] # include "LCS. h "# include <iostream> # include <fstream> using namespace std; LCS: LCS () {sX =" ABCBDAB "; sY =" BDCABA "; m = sX. size (); n = sY. size (); c = new int * [m + 1]; // the serial number of the character in the book is considered as a subscript, therefore, we need to open up a space B = new int * [m + 1]; // an empty c [0] [0] and B [0] [0] for (int I = 0; I <= m; I ++) {c [I] = New int [n + 1]; B [I] = new int [n + 1];} for (int I = 0; I <= m; I ++) {for (int j = 0; j <= n; j ++) {c [I] [j] = 0; // assign a value to B [I] [j] = 0 ;}} LCS ::~ LCS () {for (int I = 0; I <= m; I ++) {delete c [I]; // release space delete B [I];} delete c; delete B;} void LCS: ProRun () {// the time complexity of this algorithm is O (mn) for (int I = 1; I <= m; I ++) {for (int j = 1; j <= n; j ++) {if (sX [I-1] = sY [J-1]) {c [I] [j] = c [I-1] [J-1] + 1; B [I] [j] = LEFTOVER ;} else if (c [I-1] [j]> = c [I] [J-1]) {c [I] [j] = c [I-1] [j]; B [I] [j] = OVER;} else {c [I] [j] = c [I] [J-1]; B [I] [j] = LEFT ;}}} void LCS: PrintResult (int I, int j) {if (I = 0 | j = 0) {return;} if (B [I] [j] = LEFTOVER) {PrintResult (I-1, J-1); cout <sX [I-1] <endl ;} else if (B [I] [j] = OVER) {PrintResult (I-1, j);} else {PrintResult (I, J-1);} void LCS :: running () {LCS: ProRun (); LCS: PrintResult (m, n);} [cpp] # include <iostream> # include "ALSPro. h "# include" LCS. h "using namespace std; void main () {// ASL as; //. printResult (); LCS ls; ls. running ();}