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>

char S1[1001];
char S2[1001];

int LCS(int, int);

int main(){
  int n, i, commonlength;
  int len1, len2;

  scanf("%d",&n);

  for(i = 0 ; i < n ; i++){
    scanf("%s",S1);
    scanf("%s",S2);
    len1 = strlen(S1);
    len2 = strlen(S2);

    commonlength = LCS(len1,len2);
    printf("%d\n",commonlength);
  }

  return 0;
}

int LCS(int x, int y){
  int i, j;
  int C[1001][1001];

  for(i = 0 ; i <= x ; i++) C[i][0] = 0;
  for(i = 0 ; i <= y ; i++) C[0][i] = 0;

  for(i = 1 ; i <= x ; i++){
    for(j = 1 ; j <= y ; j++){
      if(S1[i-1] == S2[j-1]){
	C[i][j] = C[i-1][j-1]+1;
      }
      else if(C[i-1][j-1] >= C[i][j-1]){
	C[i][j] = C[i-1][j];
      }
      else{
	C[i][j] = C[i][j-1];
      }
    }
  }

  if(i == 0 || j == 0) return 0;
  else if(i > 0 && j > 0 && x == y) return C[i-1][j-1];
  else if(i > 0 && j > 0 && x != y) return C[x][y];
}