动态规划，求最长公共子序列

引进一个二维数组Array[][]，用Array[i][j]记录A[i]与B[j] 的LCS 的长度，sign[i][j]记录ARRAY[i][j]是通过哪一个子问题的值求得的，以决定搜索的方向。

问题的递归式写成：

 

 

回溯输出最长公共子序列过程：

// LCSLength.cpp : Defines the entry point for the console application.

//

#include "StdAfx.h"



#include "iostream"

#include "string"

#include "vector"

using namespace std;



vector< vector<int> > sign;

vector< vector<int> > array ;

string A;

string B;

void printsign(vector< vector<int> > & sign);

int getLCSlength(string strA,string strB){

    int leni = strA.length() + 1;

    int lenj = strB.length() + 1;

    int i = 0;

    int j = 0;

    // init the array

    

    array.resize(leni);

    for(i = 0;i<leni;i++)

        array[i].resize(lenj);

    // init the sign

    sign.resize(leni);

    for(i = 0;i<leni;i++)

        sign[i].resize(lenj);



    for(i = 1;i<leni;i++)

        for(j = 1;j<lenj;j++){

            if(strA[i-1] == strB[j-1]){          //dp[i][j] = dp[i-1][j-1] + 1  如果X[i-1] = Y[i-1]

                array[i][j] = array[i-1][j-1] + 1;   

                sign[i][j] = 1;      //"\"

            }

            else

                if(array[i-1][j]>=array[i][j-1]){  //下面两种情况，p[i][j] = max{ dp[i-1][j], dp[i][j-1] }  如果X[i-1] != Y[i-1]

                    array[i][j] = array[i-1][j];

                    sign[i][j] = 2;  //"^"

                }

                else{

                    array[i][j] = array[i][j-1];

                    sign[i][j] = 0;  //"<-"

                }

        }

        

        return array[leni-1][lenj-1];

}



void printsign(vector< vector<int> > & sign){

    cout<<"~~~~~~~~~~~~~~~~~~~~"<<endl;

    vector< vector<int> >::iterator iti = sign.begin();

    for(;iti!=sign.end();iti++){

        vector<int>::iterator itj = (*iti).begin();    

        for(;itj != (*iti).end();itj++){

            cout<<*itj<<" ";

        }

        cout<<endl;    

    }





}

void printLCS(int i,int j){

    

    if( i == 0 || j == 0)

        return ;

    if(sign[i][j] == 1){

        printLCS(i-1,j-1);

    //    cout<<"i = "<<i<<" j = "<<j<<" ";

        cout<<A[i-1];

    }

    if(sign[i][j] == 2){

        printLCS(i-1,j);

    }

    if(sign[i][j] == 0){

        printLCS(i,j-1);

    }

}

int main(int argc, char const *argv[])

{

    A = "ABCBDAB";

    B = "BDCABA";

    

    cout<<"the LCS length is "<<getLCSlength(A,B)<<endl;

    

    printsign(array);

    printsign(sign);

    printLCS(A.length(),B.length());

    cout<<endl;

    return 0;

}