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 Rust (judge verdict: Accepted)

use std::cmp::max;

fn find_lcs(x: &Vec<char>, y: &Vec<char>) -> u32 {
    let len_x = x.len();
    let len_y = y.len();
    let mut lengths = vec![vec![0; len_y]; len_x];

    for (i, xi) in x.iter().enumerate() {
        for (j, yj) in y.iter().enumerate() {
            if xi == yj {
                lengths[i][j] = 1;
            };
        }
    }

    // for row in &lengths {
    //     println!("{:?}", row);
    // }
    // println!();

    for (i, _) in x.iter().enumerate() {
        for (j, _) in y.iter().enumerate() {
            if i == 0 && j == 0 {
                continue;
            } else if i == 0 {
                lengths[i][j] = max(lengths[0][j - 1], lengths[i][j]);
                continue;
            } else if j == 0 {
                lengths[i][j] = max(lengths[i - 1][0], lengths[i][j]);
                continue;
            }

            lengths[i][j] = max(
                max(lengths[i - 1][j], lengths[i][j - 1]),
                lengths[i - 1][j - 1] + lengths[i][j],
            );
        }
    }

    // for row in &lengths {
    //     println!("{:?}", row);
    // }

    lengths[len_x - 1][len_y - 1]
}

fn main() {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).ok();
    let n = line.trim().parse::<usize>().unwrap();

    let mut x = Vec::new();
    let mut y = Vec::new();

    for i_line in 0..(2 * n) {
        let mut line = String::new();
        std::io::stdin().read_line(&mut line).ok();
        let inputs: Vec<char> = line.split_whitespace()
            .map(|e| e.parse::<String>().ok().unwrap())
            .collect::<String>()
            .chars()
            .collect();

        if i_line % 2 == 0 {
            x = inputs;
        } else {
            y = inputs;
            println!("{}", find_lcs(&x, &y));
        }
    }
}