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 SIZE 1000

int LCS(char*, char*, int, int);
int main(void){
    int n;
    char aString[SIZE],bString[SIZE];
  
    scanf("%d",&n);
  
    for(int i=0;i<n;i++){
        scanf("%s",aString);
        scanf("%s",bString);
        printf("%d\n",LCS(aString,bString,strlen(aString),strlen(bString)));
    }
  
    return 0;
}
  
int LCS(char *aString,char *bString,int aLength,int bLength){
    int hash[bLength+1][aLength+1];
  
    for(int i=1;i<=bLength;i++){
      hash[i][0]=0;
    }
    for(int i=0;i<=aLength;i++){
      hash[0][i]=0;
    }
  
    for(int j=1;j<=bLength;j++){
        for(int i=1;i<=aLength;i++){
            if(bString[j-1]==aString[i-1]) hash[j][i]=hash[j-1][i-1]+1;
            else if(hash[j-1][i]>=hash[j][i-1]) hash[j][i]=hash[j-1][i];
            else hash[j][i]=hash[j][i-1];
        }
    }
  
    return hash[bLength][aLength];
      
}