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 10000

int LCSLength(char *, char *);

char X[N + 1], Y[N + 1];
int c[N + 1][N + 1];

int main()
{

    int q;
    int i;
    scanf("%d", &q);

    for (i = 0; i < q; i++)
    {
        X[0] = ' ';
        Y[0] = ' ';
        scanf(" %s %s", &X[1], &Y[1]);

        // printf("%s\n%s\n",X,Y);

        printf("%d\n", LCSLength(X, Y));
    }

    return 0;
}

int LCSLength(char *X, char *Y)
{
    int i, j;
    int m, n;
    m = strlen(X);
    n = strlen(Y);

    for (i = 1; i <= m; i++)
    {
        c[i][0] = 0;
    }

    for (j = 0; j <= n; j++)
    {
        c[0][j] = 0;
    }

    for (i = 1; i <= m; i++)
    {
        for (j = 1; j <= n; j++)
        {
            if (X[i] == Y[j])
            {
                c[i][j] = c[i - 1][j - 1] + 1;
            }
            else if (c[i - 1][j] >= c[i][j - 1])
            {
                c[i][j] = c[i - 1][j];
            }
            else
                c[i][j] = c[i][j - 1];
        }
    }

    return c[m][n] - 1;
}