Longest Common subsequence (LCS)

I have never understood the LCS problem before. However, the problem of finding a job being tortured by LCs is lost with that company.

The longest common subsequence problem is a classic dynamic programming problem. The longest common subsequence problem also has the optimal substructure.

That is: xi <x1, x2 ,..., xi> that is, the first I characters (1 <= I <= m), YJ <Y1, Y2 ,..., YJ) that is, the first J sequences of the Y sequence (1 <= j <= N); assume z <Z1, Z2 ,..., ZK> belongs to LCS (x, y );

If: XM = yn (the last character is the same), it is not difficult to prove that this character must be any of the longest common subsequences of X and y z (set length to K) the last character, that is, zk = XM = YN and apparently there is a Zk-1 in LCS (Xm-1, Yn-1) That is Z prefix Zk-1 is the longest common subsequence of Xm-1 and Yn-1. At this time, the question is to find the Xm-1 and Yn-1 LCS (The length of LCS (X, Y) is equal to that of LCS (Xm-1, Yn-1).

If: XM = YN, it is not difficult to use reverse proof: either Z, LCS (Xm-1, Y), or Z, LCS (x, Yn-1 ). Because ZK and ZK have at least one of them must be true, ZK and XM has Z, LCS (Xm-1, Y), similar, if ZK is not YN, there is Z in LCS (x, Yn-1 ). At this time, the question is to find the Xm-1 and Y LCs and X and Yn-1 LCS. The length of LCS (X, Y) is: the length of max {LCS (Xm-1, Y), the length of LCS (x, Yn-1 }.

Since the above when XM =yn, finding the length of LCS (Xm-1, Y) and LCS (x, Yn-1) are not mutually independent of each other: both require the length of LCS (Xm-1, Yn-1. The LCS of the other two sequences contains the LCS of the prefix of the two sequences. Therefore, the dynamic programming method is used for the problem of optimal sub-structure.

That is to say, to solve this LCS problem, you require three things: 1. LCS (Xm-1, Yn-1) + 1; 2. LCS (Xm-1, Y), LCS (X, yn-1); 3, max {LCS (Xm-1, Y), LCS (x, Yn-1 )}.

The recursive formula can be obtained from the above:

 

#include <iostream>#include <vector>#include <cstring>using namespace std;enum decreaseDir {kInit = 0, kLeft, kUp, kLeftUp};string ret;void LCS_Print(vector<vector<int> > &dir, const char *str1, const char *str2, int row, int col){    if(str1 == NULL || str2 == NULL)        return;    int len1 = strlen(str1);    int len2 = strlen(str2);    if(len1 == 0 || len2 == 0 || !(row < len1 && col < len2 ))        return;    if(dir[row][col] == kLeftUp)    {        if(row > 0 && col > 0)            LCS_Print(dir, str1, str2, row - 1, col -1);        ret.push_back(str1[row]);    }    else if(dir[row][col] == kUp)    {        if(row > 0)            LCS_Print(dir, str1, str2, row - 1, col);    }    else if(dir[row][col] == kLeft)    {        if(col > 0)            LCS_Print(dir, str1, str2, row, col -1);    }}int LCS(const char *str1, const char *str2){    if(!str1 || !str2)        return 0;    int len1 = strlen(str1);    int len2 = strlen(str2);    if(!len1 || !len2)        return 0;    int i, j;    vector<vector<int> > dp(len1, vector<int>(len2, 0));    vector<vector<int> > dir(len1, vector<int>(len2, 0));    for(int i = 0; i < len1; i++)    {        for(int j = 0; j < len2; j++)        {            if(i == 0 || j == 0)            {                if(str1[i] == str2[j])                {                    dp[i][j] = 1;                    dir[i][j] = kLeftUp;                }                else                {                    if(i > 0)                    {                        dp[i][j] = dp[i-1][j];                        dir[i][j] = kUp;                    }                    if(j > 0)                    {                        dp[i][j] = dp[i][j-1];                        dir[i][j] = kLeft;                    }                }            }            else if(str1[i] == str2[j])            {                dp[i][j] = dp[i-1][j-1] + 1;                dir[i][j] = kLeftUp;            }            else if(dp[i-1][j] > dp[i][j-1])            {                dp[i][j] = dp[i-1][j];                dir[i][j] = kUp;            }            else            {                dp[i][j] = dp[i][j-1];                dir[i][j] = kLeft;            }        }    }    LCS_Print(dir, str1, str2, len1 - 1, len2 - 1);    return dp[len1-1][len2-1];}int main(){    const char *str1 = "abcde";    const char *str2 = "acde";    ret.clear();    int longest = LCS(str1, str2);    cout << "longest = " << longest << endl;    cout << "ret = " << ret << endl;    return 0;}

 

Longest Common subsequence (LCS)