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 C++ (judge verdict: Accepted)

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//#endif
#include<iostream>
#include<string>
#define lth 1010

using namespace std;

int main()
{
	int n, L[1010][1010], Lmax = 0, *save;
	string *Xi, *Yi;
	cin >> n;
	save = new int[n];
	if (n <= 20)
	{
		Xi = new string[lth];
		Yi = new string[lth];
	}
	else
	{
		Xi = new string[1010];
		Yi = new string[1010];
	}
	for (int k = 0; k < n; k++)
	{
		Lmax = 0;
		string X, Y;
		cin >> X >> Y;
		for (int i = 0; i <= X.length(); i++)
		{
			if (i != 0)
				Xi[i] = X.substr(0, i );
		}
		for (int i = 0; i <= Y.length(); i++)
		{
			if (i != 0)
				Yi[i] = Y.substr(0, i );
		}
		for (int i = 0; i <= X.length(); i++)
			for (int j = 0; j <= Y.length(); j++)
			{
				if (i == 0 || j == 0)
					L[i][j] = 0;
				else if (Xi[i][i - 1] == Yi[j][j - 1])
				{
					L[i][j] = L[i - 1][j - 1] + 1;
				}
				else
				{
					L[i][j] = (L[i][j - 1] > L[i - 1][j]?L[i][j - 1] : L[i - 1][j]);
				}
				if (Lmax < L[i][j]) Lmax = L[i][j];
			}
		save[k] = Lmax;
	}

	for (int i = 0; i < n; i++)
		cout << save[i] << endl;
}