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 JAVA (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
    public static int LCS(String s1, String s2) {
        int[][] matrix = new int[s1.length() + 1][s2.length() + 1];
        for (int i = 0; i < matrix.length; i++) {
            matrix[i][0] = 0;
        }

        for (int i = 0; i < matrix[0].length; i++) {
            matrix[0][i] = 0;
        }

        for (int i = 1; i < matrix.length; i++) {
            for (int j = 1; j < matrix[0].length; j++) {
                matrix[i][j] = s1.charAt(i - 1) == s2.charAt(j - 1) ? 1 + matrix[i - 1][j - 1] : Math.max(matrix[i - 1][j], matrix[i][j - 1]);
            }
        }
        return matrix[s1.length()][s2.length()];
    }
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int queries = input.nextInt();
        input.nextLine();

        for (int i = 0; i < queries; i++) {
            System.out.println(LCS(input.nextLine(), input.nextLine()));
        }

        input.close();
    }
}