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 MAX 1000

void LCSLength(char *X, char *Y){

    int i,j;
    int m = strlen(X);
    int n = strlen(Y);
    int c[m+1][n+1];

    for( i = 0 ; 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-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];
            }
        }
    }
    

    printf("%d\n",c[i-1][j-1]);

    
}



int main(){

    int n;
    int i;
    char line1[MAX],line2[MAX];

    scanf("%d",&n);

    for( i = 0 ; i < n ; i++ ){
        scanf("%s",line1);
        scanf("%s",line2);
        LCSLength(line1,line2);
    }

    return 0;
}