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>

int dynamic(char X[], char Y[]) {
    int c[1001][1001], i, j, x, y, a;
    a = 0;

    x = strlen(X);
    y = strlen(Y);
    
    for (i = 0; i <= x; ++i) c[i][0] = 0;
    for (j = 0; j <= y; ++j) c[0][j] = 0;
    
    for (i = 1; i <= x; ++i) {
        for (j = 1; j <= y; ++j) {
            if (X[i-1] == Y[j-1]) {
                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];
                }
            }
            if (a < c[i][j]) a = c[i][j];
        }
    }
    return a;
}


int main() {
    int q, a, k, t[150];
    char X[1000], Y[1000];
    a = 0;
    scanf("%d", &q);
    
    for (k = 0; k < q; ++k) {
        scanf("%s", X);
        scanf("%s", Y);
        t[k] = dynamic(X, Y);
    }
    
    for (k = 0; k < q; ++k) {
        printf("%d\n", t[k]);
    }
    return 0;
}