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)

/**
 * Dynamic Programming - Longest Common Subsequence
 */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

import static java.lang.Integer.parseInt;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		int q = parseInt(br.readLine());
		for (int i = 0; i < q; i++) {
			String X = br.readLine();
			String Y = br.readLine();

			int[][] dp = new int[X.length() + 1][Y.length() + 1];
			int lcs = 0;
			for (int j = 1; j <= X.length(); j++) {
				for (int k = 1; k <= Y.length(); k++) {
					char x = X.charAt(j - 1);
					char y = Y.charAt(k - 1);
					if (x == y) {
						lcs = dp[j][k] = dp[j - 1][k - 1] + 1;
					} else {
						lcs = dp[j][k] = Math.max(dp[j][k - 1], dp[j - 1][k]);
					}
				}
			}
			System.out.println(lcs);
		}
	} //end main
}