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

#define MAX(a,b) (((a)>(b))?(a):(b))

#define MAX_LENGTH 1000 + 1

int q;

int main() {
	scanf("%d", &q);
	int numInput = 2 * q;

	int lcs;

	char **input = (char**)malloc(numInput * sizeof(char*));
	for (int i = 0; i < numInput; i++) input[i] = (char*)malloc(MAX_LENGTH * sizeof(char));
	for (int i = 0; i < numInput; i++) {
		scanf("%s", input[i]);
		if (i % 2 != 0) {
			lcs = LCSLength(input[i - 1], input[i]);
			printf("%d\n", lcs);
		}
	}

	for (int i = 0; i < numInput; i++) free(input[i]);
	free(input);

	return 0;
}

int LCSLength(char A[], char B[]) {
	int m = strlen(A) + 1;
	int n = strlen(B) + 1;

	int maxLCS = 0;

	int** C = (int**)malloc(m * sizeof(int*));
	for (int i = 0; i < m; i++) C[i] = (int*)calloc(n, sizeof(int));

	for (int i = 1; i < m; i++) {
		for (int j = 1; j < n; j++) {
			if (A[i-1] == B[j-1]) C[i][j] = 1 + C[i - 1][j - 1];
			else C[i][j] = MAX(C[i - 1][j], C[i][j - 1]);
			maxLCS = MAX(C[i][j], maxLCS);
		}
	}

	return maxLCS;
}