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 N 1005


main(){
  
  int n,i;
  char X[N],Y[N];
  
  
  
  scanf("%d",&n);
  
  for(i = 0; i < n; i++){
    scanf("%s%s",X+1,Y+1);
    Lcs(X,Y);
  }
  
  return 0;
  
}
int Lcs(char *X,char *Y)
{
  
  int m,n,i,j;
  int Z[N][N];
  
  m = strlen(X+1);
  n = strlen(Y+1);

  for(i = 0; i <= m; i++){
    Z[i][0] = 0;
  }
  for(j = 0; j <= n; j++){
    Z[0][j] = 0;
  }
  
  for(i = 1; i <= m; i++){
    for(j = 1; j <= n; j++){
      
      if(X[i] == Y[j]){
	Z[i][j] = Z[i-1][j-1] +1;
      }
      else if(Z[i-1][j] >= Z[i][j-1]){
	Z[i][j] = Z[i-1][j];
      }
      else {
	Z[i][j] = Z[i][j-1];
      }
    }
  }
  
  
  printf("%d\n",Z[i-1][j-1]);
}