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<stdio.h>
#include<string.h>
#define N 1000

int Max(int a, int b){
	if(a >= b){
		return a;
	} else {
		return b;
	}
}

int lcs(char x[], char y[]){
	int i, j;
	int c[N+1][N+1];
	int m = strlen(x);
	int n = strlen(y);
	int max = 0;
	
	char x1[m+1] , y1[n+1];
	
	x1[0] = ' ';
	for(i = 1; i <= m; i++ ){
		x1[i]=x[i-1];
	}
	y1[0] = ' ';
	for(i = 1; i <= n; i++ ){
		y1[i]=y[i-1];
	}
	
	for(i = 0; i <= m; i++){
		c[i][0] = 0;
	}
	
	for(j = 1; j <= n; j++){
		c[0][j] = 0;
	}
	
	for(i = 1; i <= m; i++){
		for(j = 1; j <= n; j++){
			if(x1[i]==y1[j]){
				c[i][j] = c[i-1][j-1] + 1;
			} else {
				c[i][j] = Max(c[i-1][j], c[i][j-1]);
			}
			max = Max(max, c[i][j]);
		}
	}
	
	return max;
}

int main( void ){
	
	char s1[N] , s2[N];
	
	int n;
	
	scanf( "%d" , &n );
	
	for( int i = 0; i < n; i++ ){
		scanf( "%s" , s1 );
		scanf( "%s" , s2 );
		printf( "%d\n" , lcs(s1 , s2) );
	}
	
	return 0;
}