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 <string.h>

#define N 1000
#define max(a,b) ( a > b ? a : b )

int main(void)
{
    int iii, i, j, m, n;
    char x[N], y[N];
    int table[N+1][N+1];
    int num;
    scanf( "%d", &num );

    for( iii=0; iii<num; ++iii ) {
        scanf( "%s", x );
        scanf( "%s", y );

        m = strlen(x);
        n = strlen(y);

        for( i=0; i<=m; ++i )
            for( j=0; j<=n; ++j )
                table[i][j] = 0;

        for( i=1; i<=m; ++i )
            for( j=1; j<=n; ++j )
                if ( x[i-1] == y[j-1] )
                    table[i][j] = table[i-1][j-1] + 1;
                else
                    table[i][j] = max( table[i][j-1], table[i-1][j] );

        printf( "%d\n", table[m][n] );
    }

    return 0;
}