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 C++ (judge verdict: Accepted)

#include<iostream>
#include<iomanip>
#include<algorithm>
#include<bitset>
#include<cctype>
#include<climits>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<deque>
#include<list>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<sstream>
#include<queue>
#include<vector>
using namespace std;

#define MAX 1001

static int dp[MAX][MAX];

int main() {
	int q; cin >> q;
	while (q--) {
		memset(dp, 0, sizeof(dp));
		string X, Y; cin >> X >> Y;
		for (int i = 1; i <= X.size(); i++) {
			for (int j = 1; j <= Y.size(); j++) {
				if (X[i - 1] == Y[j - 1]) {
					//cout << i << j << X[i - 1] << endl;
					dp[i][j] = dp[i - 1][j - 1] + 1;
				}
				else {
					dp[i][j] = max(
						dp[i - 1][j],
						dp[i][j - 1]
					);
				}
			}
		}
		//for (int i = 0; i <= Y.size(); i++) {
		//	for (int j = 0; j <= X.size(); j++) {
		//		cout << dp[j][i] << " ";
		//	}
		//	cout << endl;
		//}
		cout << dp[X.size()][Y.size()] << endl;

	}

	return 0;
}