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>

int LCS(char X[1000] , char Y[1000]) {
    int Z[1010][1010];
    int MAX = 0;
    int i , j;
    
    int Len_X = strlen(X);
    int Len_Y = strlen(Y);
    
    for(i = 0 ; i <= Len_X ; i++)
        Z[i][0] = 0;
    for(j = 0 ; j <= Len_Y ; j++)
        Z[0][j] = 0;
    
    for(i = 1 ; i <= Len_X ; i++) {
        for(j = 1 ; j <= Len_Y ; j++){
            if(X[i - 1] == Y[j - 1])
                Z[i][j] = Z[i - 1][j - 1] + 1;
            else
                Z[i][j] = (Z[i - 1][j] > Z[i][j - 1]) ? Z[i - 1][j] : Z[i][j - 1];
            
            MAX = (MAX > Z[i][j]) ? MAX : Z[i][j];
        }
    }
    
    return MAX;
}

int main(void)
{
    int q , i;
    
    scanf("%d" , &q);
    
    char X[1000] , Y[1000];
    
    for (i = 0 ; i < q ; i++) {
        scanf("%s" ,X);
        scanf("%s" ,Y);
        
        printf("%d\n" , LCS(X , Y));
    }
    
    return 0;
}