Dynamic Planning-Longest Common subsequence (LCS)

----- Edit by ZhuSenlin HDU

Returns the longest common subsequence. Given two sequences and
To find the largest common subsequences of X and Y.

1) determine the optimal sub-structure of LCS. If any LCS is set to X and Y

If xm = yn, then zk = xm = yn and the Zk-1 is an lcs of the Xm-1 and Yn-1;

If xm! = Yn, so zk! = Xm contains Z is an LCS of Xm-1 and Y;

If xm! = Yn, so zk! = Yn contains Z is an LCS of X and Yn-1

2) Search for Recursive Solutions

Set C [I, j] to the length of an LCS of the sequence Xi and Yj. Then

3) Calculate the length

Create a Table to store the length of each (I, j) LCS. Then C [Length (X)] [Length (Y)] is the LCS Length of X and Y.

Create another table Path record Path. If Path [I] [j] = 'X' (oblique ), if Path [I] [j] = 'U' (up), if Path [I] [j] = 'l' (left ).

 

The Code is as follows:

template <class T>void CreateTable(T**& pTable, int nH, int nW){DeleteTable(pTable,nH);pTable = new T* [nH];for(int i = 0; i < nH; i++){pTable[i] = new T[nW];memset(pTable[i],0,nW*sizeof(T));}}template <class T>void DeleteTable(T**& pTable, int nH){if(!pTable)return;for(int i = 0; i < nH; i++)if(pTable[i])delete pTable[i];delete pTable;}int LengthLCS(const char* pStrA, const char* pStrB, int**& pTable, char**& pPath){int nLengthA = strlen(pStrA);int nLengthB = strlen(pStrB);for(int i = 1; i<= nLengthA; i++){for(int j = 1; j <= nLengthB; j++) {if(pStrA[i-1] == pStrB[j-1]) {pTable[i][j] = pTable[i-1][j-1]+1;pPath[i][j] = 'x';}else if(pTable[i-1][j] < pTable[i][j-1]){pTable[i][j] = pTable[i][j-1];pPath[i][j] = 'l';}else {pTable[i][j] = pTable[i-1][j];pPath[i][j] = 'u';}}}return pTable[nLengthA][nLengthB];}void PrintLCS(char** pPath, const char* pStrA,int nLengthA, int nLengthB){if(nLengthA == 0 || nLengthB == 0)return;if(pPath[nLengthA][nLengthB] == 'x') {PrintLCS(pPath, pStrA, nLengthA-1, nLengthB-1);cout << pStrA[nLengthA-1] << " ";}else if(pPath[nLengthA][nLengthB] == 'l')PrintLCS(pPath,pStrA,nLengthA,nLengthB-1);elsePrintLCS(pPath,pStrA,nLengthA-1,nLengthB);}

The test code is as follows:

int main(){const char* pStrA = "abcbdab";const char* pStrB = "bdcaba";int** pTable = NULL;char** pPath = NULL;int nLengthA = strlen(pStrA);int nLengthB = strlen(pStrB);CreateTable(pTable,nLengthA+1,nLengthB+1);CreateTable(pPath,nLengthA+1,nLengthB+1);cout << LengthLCS(pStrA,pStrB,pTable,pPath) << endl;PrintLCS(pPath,pStrA,strlen(pStrA),strlen(pStrB));cout << endl;DeleteTable(pTable,strlen(pStrA)+1);DeleteTable(pPath,strlen(pStrA)+1);return 0;}

 

Reference books: Introduction to Algorithms