Problem:
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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.*;
import java.io.*;

/**
 * C
 */
public class Main {

    void run() {
        
        Scanner sc = new Scanner(System.in);
        int q = Integer.parseInt(sc.nextLine());

        for(int i=0; i<q; i++) {
            System.out.println(ans(sc.nextLine(), sc.nextLine()));
        }

    }

    int ans(String x, String y) {
        
        int[][] DP = new int[x.length()+1][y.length()+1];

        for(int i=1; i<=x.length(); i++) {
            for(int j=1; j<=y.length(); j++) {
                
                DP[i][j] = Math.max(DP[i][j-1], DP[i-1][j]);

                if(x.charAt(i-1) == y.charAt(j-1)) {
                    DP[i][j] = Math.max(DP[i][j], DP[i-1][j-1]+1);
                }

            }
        }

        return DP[x.length()][y.length()];

    }

    public static void main(String[] args) {
        new Main().run();
    }

}