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.io.IOException;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) throws IOException {

		Scanner in = new Scanner(System.in);
		int q = in.nextInt();
		char[][] x = new char[q][];
		char[][] y = new char[q][];

		for (int i = 0; i < q; i++) {
			x[i] = in.next().toCharArray();
			y[i] = in.next().toCharArray();
		}

		for (int i = 0; i < q; i++) {
			// dp
			int[][] dp = new int[x[i].length+1][y[i].length+1];
			for (int j = 0; j < x[i].length; j++) {
				for (int l = 0; l < y[i].length; l++) {
					if (x[i][j] == y[i][l]) {
						dp[j+1][l+1] = dp[j][l]+1;
					} else {
						dp[j+1][l+1] = Math.max(dp[j+1][l], dp[j][l+1]);
					}
				}
			}
			System.out.println(dp[x[i].length][y[i].length]);
		}

	}
}