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>
#define MAXLEN 1001

char x[MAXLEN];
char y[MAXLEN];
int  b[MAXLEN][MAXLEN];                  //传递二维数组必须知道列数，所以使用MAXLEN这个确定的数


int max(int a,int c){
    if(a>c) return a;
    else return c;
}

int LCS()
{
    int i, j;
    int m = strlen(y);
    int n = strlen(x);
    memset(b,0,sizeof b);
    int maxx = 0;
    for(i=0;i<n;i++) {
        for(j=0;j<m;j++) {
            if(i!=0) b[i][j] = max(b[i][j],b[i-1][j]);
            if(j!=0) b[i][j] = max(b[i][j],b[i][j-1]);
            if(x[i] == y[j]) {
                b[i+1][j+1] = max(b[i+1][j+1],b[i][j]+1);
            maxx = max(maxx,b[i+1][j+1]);
            }
        }
    }
    return maxx;
}

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