August 26-Dynamic Planning of LCS

Solve the longest public subsequence problem:

Solution:

For example, the given two sequences are x = <A, B, C, B, D, a, B> and Y = <B, D, C, A, B, a>. The results calculated by the algorithms lcs_length and LCs are shown in:

The template can be written

void lcss(){    int i,j;    int sizex=str1.length();    int sizey=str2.length();    for(i=0;i<=sizex;i++)        lcs[i][0]=0;    for(i=0;i<=sizey;i++)         lcs[0][i]=0;        for(i=1;i<=sizex;i++)            for(j=1;j<=sizey;j++)            {                if(str1[i-1]==str2[j-1])                    lcs[i][j]=lcs[i-1][j-1]+1;                else                    lcs[i][j]=lcs[i-1][j]>=lcs[i][j-1]?lcs[i-1][j]:lcs[i][j-1];            }                          cout<<lcs[sizex][sizey]<<endl;  }  

View code

You can also compress the array:

memset(lcs,0,sizeof(lcs));        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)            {                if(str1[i-1]==str2[j-1])                    lcs[i%2][j]=lcs[(i-1)%2][j-1]+1;                else                    lcs[i%2][j]=lcs[(i-1)%2][j]>lcs[i%2][j-1]?lcs[(i-1)%2][j]:lcs[i%2][j-1];            }    printf("%d\n",lcs[n%2][n]) ; 

View code

 

 

Training Questions:

Http://acm.hdu.edu.cn/diy/contest_show.php? Cid = 24575

August 26-Dynamic Planning of LCS