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>
int xy[1001][1001];
char str1[1001];
char str2[1001];
int main(){
  int q;
  int i;
  int j,k;
  scanf("%d",&q);
  for(i=0;i<q;i++){
    scanf("%s%s",str1,str2);
    for(j=1;j<=(int)strlen(str1);j++){
      for(k=1;k<=(int)strlen(str2);k++){
	if(str1[j-1]==str2[k-1]){
	  xy[j][k] = xy[j-1][k-1] + 1;
	}
	else{
	  xy[j][k] = xy[j-1][k] > xy[j][k-1] ? xy[j-1][k] : xy[j][k-1];
	}
      }
    }
    printf("%d\n",xy[strlen(str1)][strlen(str2)]);
    /*
    for(j=0;j<1001;j++)
      for(k=0;k<1001;k++) xy[j][k]=0;
    */
  }
  
  return 0;
}