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 <string>
#include <vector>
#include <set>
#include<algorithm>
#include <iomanip>
int ctoi(char c) {
	switch (c) {
	case '0': return 0;
	case '1': return 1;
	case '2': return 2;
	case '3': return 3;
	case '4': return 4;
	case '5': return 5;
	case '6': return 6;
	case '7': return 7;
	case '8': return 8;
	case '9': return 9;
	default: return 0;
	}
}
using namespace std;
typedef long long ll;
//定義場所//
int i, j, k, l, m, n;
int N, Q;
vector <int> Ans;
////////////

int main()
{
	int N;
	cin >> N;
	for (k = 0; k < N; k++) {
		vector<vector<int> >dp(1001, vector<int>(1001, 0));
		int ans = 0;
		string s1, s2;
		cin >> s1 >> s2;
		s1 = ' ' + s1;
		s2 = ' ' + s2;
		for (i = 0; i < s1.size(); i++) { dp[i][0] = 0; }
		for (i = 1; i < s2.size(); i++) { dp[0][i] = 0; }
		for (i = 1; i < s1.size(); i++) {
			for (j = 1; j < s2.size(); j++) {
				if (s1[i] == s2[j]) {
					dp[i][j] = dp[i - 1][j - 1] + 1;
				}
				else {
					dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
				}
				ans = max(ans, dp[i][j]);
			}
		}
		Ans.push_back(ans);
	}
	for (i = 0; i < Ans.size(); i++) {
		cout << Ans[i] << endl;
	}
}