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;
use std::io::stdin;
fn main() {
    let q: i32 = read();
    let xy = (0..q).map(|_| (read::<String>(), read::<String>()));
    for (x, y) in xy {
        println!("{}", lcs(x.as_str(), y.as_str()));
    }
}

fn lcs(x: &str, y: &str) -> i32 {
    let x: Vec<char> = x.chars().collect();
    let y: Vec<char> = " ".chars().chain(y.chars()).collect();
    let mut dp: Vec<Vec<i32>> = vec![vec![0; y.len() + 1]; x.len() + 1];
    for i in 0..x.len() {
        for j in 0..y.len() {
            dp[i + 1][j + 1] = if x[i] == y[j] {
                max(dp[i][j] + 1, max(dp[i][j + 1], dp[i + 1][j]))
            } else {
                max(dp[i + 1][j], dp[i][j + 1])
            }
        }
    }
    dp[x.len()][y.len()]
}

fn read<T: std::str::FromStr>() -> T {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim().parse().ok().unwrap()
}