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;
}