Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

Longest Common Subsequence

For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \{a,b,c,b,d,a,b\}$ and $Y = \{b,d,c,a,b,a\}$, the sequence $\{b,c,a\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\{b,c,a\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\{b,c,b,a\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\{b,c,b,a\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater.

Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.

Input

The input consists of multiple datasets. In the first line, an integer $q$ which is the number of datasets is given. In the following $2 \times q$ lines, each dataset which consists of the two sequences $X$ and $Y$ are given.

Output

For each dataset, print the length of LCS of $X$ and $Y$ in a line.

Constraints

$1 \leq q \leq 150$

$1 \leq$ length of $X$ and $Y$ $\leq 1,000$

$q \leq 20$ if the dataset includes a sequence whose length is more than 100

Sample Input 1

3
abcbdab
bdcaba
abc
abc
abc
bc

Sample Output 1

4
3
2

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

#define MAX 1000

int getlcs(char *X, char *Y,int m,int n)
{
  int DATA[MAX+1][MAX+1];
  int maxlcs;

  for(int i=0;i<m;i++) DATA[i][0]=0;
  for(int j=1;j<n;j++) DATA[0][j]=0;

  for(int i=1;i<=m;i++)
  {
    for(int j=1;j<=n;j++)
    {
      if(X[i]==Y[j]) DATA[i][j]=DATA[i-1][j-1]+1;
      else if(DATA[i-1][j]>DATA[i][j-1]) DATA[i][j]=DATA[i-1][j];
      else DATA[i][j]=DATA[i][j-1];
      maxlcs=DATA[i][j];
    }
  }

  return maxlcs;
}

int main()
{
  char s1[MAX];
  char s2[MAX];
  int n;
  char change;
  int s1_l,s2_l;

  scanf("%d",&n);
  for(int i=0;i<n;i++)
  {
    scanf("%s %s", s1,s2);
    s1_l=strlen(s1);
    s2_l=strlen(s2);

    for(int j=s1_l;j>0;j--)
    {
      s1[j]=s1[j-1];
    }
    s1[0]=' ';

    for(int j=s2_l;j>0;j--)
    {
      s2[j]=s2[j-1];
    }
    s2[0]=' ';
    printf("%d\n",getlcs(s1,s2,s1_l,s2_l));
  }
  return 0;
}