This program in C finds 2 Longest common subsequence of 2 given strings(say X and Y),entered without any space on the screen individually,altering X and Y strings.I compiled and executed this above program successfully in Dev Cpp compiler as a C file(not a c++ file).

#include<stdio.h>
#include<conio.h>
#include<string.h>
void print_lcs(char b[][20],char x[],int i,int j)
{
     if(i==0 || j==0)
      return;
     if(b[i][j]=='c')
      {
       print_lcs(b,x,i-1,j-1);
       printf(" %c",x[i-1]);
       }
      else if(b[i][j]=='u')
       print_lcs(b,x,i-1,j);
      else
       print_lcs(b,x,i,j-1);
}  
void lcs_length(char x[],char y[])
{
     int m,n,i,j,c[20][20];
     char b[20][20];
     m=strlen(x);
     n=strlen(y);    
     for(i=0;i<=m;i++)
      c[i][0]=0;
     for(i=0;i<=n;i++)
      c[0][i]=0;
     for(i=1;i<=m;i++)
      for(j=1;j<=n;j++)
      {
	if(x[i-1]==y[j-1])
	 {
	   c[i][j]=c[i-1][j-1]+1; 
	   b[i][j]='c';           \\c stands for left upright cross
	   }
	 else if(c[i-1][j]>=c[i][j-1])
	 {
	      c[i][j]=c[i-1][j];
	      b[i][j]='u';         \\u stands for upright or above
	      }
	 else
	 {
	     c[i][j]=c[i][j-1];
	     b[i][j]='l';         \\l stands for left
	     }
	}          
    print_lcs(b,x,m,n);      
 }     
void lcs()
{
     int i,j;
     char x[20],y[20];     
     printf("1st sequence:");
     gets(x);
     printf("2nd sequence:");
     gets(y); 
     printf("\nlcs are:");
     lcs_length(x,y); 
     printf("\n");
     lcs_length(y,x);   
 }
main()
{
     char ch;
     do
     {
           lcs();
           printf("\nContinue(y/n):");
           ch=getch();
           }
     while(ch=='y'||ch=='Y');
}

Hi, could you please briefly describe the code? Like, which variable represent what (eg b, i, j...) and print_lcs, lcs_lenght - how to they work? i am confused and lost, thank a lot for a help....

#include<stdio.h>
#include<conio.h>
void print_lcs(char b[10][10],int x[10],int i,int j)
{
    if(i==0||j==0)
    {
        return;
    }
    if(b[i][j]=='d')
    {
        print_lcs(b,x,i-1,j-1);
        printf("%d",x[i-1]);
    }
    else
    {
        if(b[i][j]=='u')
        {
            print_lcs(b,x,i-1,j);
        }
        else
        {
            print_lcs(b,x,i,j-1);
        }
    }
}
void main()
{
    int m,n,i,j,c[10][10],x[10],y[10];
    char b[10][10];
    clrscr();
    printf("\n Enter the length of first array: ");
    scanf("%d",&m);
    printf("\n Enter %d elements: ",m);
    for(i=0;i<m;i++)
    {
        scanf("%d",&x[i]);
    }
    printf("\n Enter the length of second array: ");
    scanf("%d",&n);
    printf("\n Enter %d elements :",n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&y[i]);
    }
    for(i=0;i<=m;i++)
    {
        c[i][0]=0;
    }
    for(j=0;j<=n;j++)
    {
        c[0][j]=0;
    }
    for(i=1;i<=m;i++)
    {
        for(j=1;j<=n;j++)
        {
            if(x[i-1]==y[j-1])
            {
                c[i][j]=c[i-1][j-1]+1;
                b[i][j]='d';
            }
            else
            {
                if(c[i-1][j]>=c[i][j-1])
                {
                    c[i][j]=c[i-1][j];
                    b[i][j]='u';
                }
                else
                {
                    c[i][j]=c[i][j-1];
                    b[i][j]='s';
                }
            }
        }
    }
    printf("\n Matrix C is:\n");
    for(i=0;i<=m;i++)
    {
        for(j=0;j<=n;j++)
        {
            printf("%d ",c[i][j]);
        }
        printf("\n");
    }
    printf("\n Matrix B is:\n");
    for(i=1;i<=m;i++)
    {
        for(j=1;j<=n;j++)
        {
            printf("%c ",b[i][j]);
        }
        printf("\n");
    }
    printf("\n The longest common subsequence is: ");
    print_lcs(b,x,m,n);
    getch();
}

m=holds length of first sequence.
n=holds length of second sequence.
x=name of first sequence array.
y=name of second sequence array.
c=a matrix of length of an LCS of the sequence x and y
If either i=0 or j=0, one of the sequences has length 0, so the LCS has length 0.
The optimal substructure of the LCS problem gives the recursive formula

c[i,j]= 0 ,if i=0 or j=0
= c[i-1,j-1]+1 ,if i,j>0 and x[ i ] = y[ j ]
= max( c[ i , j-1] , c[ i-1 , j]) ,if i,j>0 and x[ i ] != y[ j ]

It also maintains the table b[1..m , 1..n] to simplify construction of an optimal solution. Intuitively,
b[ i , j ] points to the table entry corresponding to the optimal subproblem solution chosen when compting c[ i , j ].
The procedure returns the b and c tables.
c[m , n] contains the length of an LCS of X and Y.

The function "print_lcs" is used to print the longest common subsequence pattern