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>
#include<stdlib.h>

int i;
char x[1001],y[1001];
int max(int,int);
int main(){
    int i,n;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%s %s",x,y);
        printf("%d\n",lcs(x,y));
    }
    return 0;
}

int lcs(){
    int j,m,n;
    int sum=0;
    int c[1001][1001];
    m = strlen(x);
    n = strlen(y);
    for(i = 0;i < m + 1;i++) c[i][0]=0;
    for(j = 0;j < n + 1;j++) c[0][j]=0;
    for(i = 1;i < m + 1;i++){
        for(j = 1;j < n + 1;j++){
            if(x[i-1] == y[j-1]) c[i][j] = c[i-1][j-1] + 1;
            else c[i][j] = max(c[i-1][j],c[i][j-1]);
            sum = max(sum,c[i][j]);
        }
    }
    return sum;
}

int max(int l,int r){
    if(l<r)return r;
    else return l;
}