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 <stdio.h>
#include <stdlib.h>
#include <string.h>

int getLCS(char*, char*);

main(){
	char x[1001], y[1001];
	int q, i;

	scanf("%d", &q);
	for (i = 0; i < q; i++){
		scanf("%s%s", x, y);
		printf("%d\n", getLCS(x, y));
	}
	return 0;
}

int getLCS(char *x, char *y){
	int i, j, lenX, lenY;
	static int C[1001][1001]= {{0}};

	lenX = strlen(x);
	lenY = strlen(y);

	for (i = 1; i <= lenX; i++){
		for (j = 1; j <= lenY; j++){
			if (x[i-1] == y[j-1]) C[i][j] = C[i-1][j-1] + 1;
			else C[i][j] = ((C[i-1][j] >=  C[i][j-1])?  C[i-1][j] : C[i][j-1]);
		}
	}

	return C[lenX][lenY];
}