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)


#include <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <string>

using namespace std;

int q;
int main() {
	cin >> q;
	for (int i = 0; i < q; i++) {
		string x, y;
		cin >> x >> y;
		int m = x.size();
		int n = y.size();
		x = ' ' + x;
		y = ' ' + y;
		int maxl = 0;
		int c[1001][1001];
		c[0][0] = 0;
		for (int i = 1; i <= m; i++) {
			c[i][0] = 0;
		}
		for (int j = 1; j <= n; j++) {
			c[0][j] = 0;
		}
		for (int i = 1; i <= m; i++) {
			for (int j = 1; j <= n; j++) {
				if (x[i] == y[j]) {
					c[i][j] = c[i-1][j-1] + 1;
				} else {
					c[i][j] = max(c[i][j - 1], c[i - 1][j]);
				}
				maxl = max(maxl, c[i][j]);
			}
		}
		cout << maxl << endl;
	}
	return 0;
}