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<algorithm>
using namespace std;

string S,T;
int x[10000],y[10000];
int dp[10000][10000];
char U[53]="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
int MAXN;

int main(){
	int q;
	cin>>q;
	for(int i=0;i<q;i++){
		MAXN=0;
		cin>>S>>T;
		for(int i=0;i<S.size();i++){
			for(int j=0;j<52;j++){
				if(S[i]==U[j]){
					x[i+1]=j;
				}
			}
		}
		for(int i=0;i<T.size();i++){
			for(int j=0;j<52;j++){
				if(T[i]==U[j]){
					y[i+1]=j;
				}
			}
		}
		for(int i=0;i<=T.size();i++){
			for(int j=0;j<=S.size();j++){
				if(i==0 || j==0){
					dp[i][j]=0;
				}
				else if(x[j]==y[i]){
					dp[i][j]=dp[i-1][j-1]+1;
				}
				else{
					dp[i][j]=max(dp[i][j-1],dp[i-1][j]);
				}
			}
		}
		for(int i=0;i<=T.size();i++){
			for(int j=0;j<=S.size();j++){
				MAXN=max(MAXN,dp[i][j]);
			}
		}
		cout<<MAXN<<endl;
	}
	return 0;
}