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 <stdlib.h>
#include <string.h>
#define N 1000

int max(int,int);

int main(){
  int q,i,j,k,lenA,lenB,count[N+1][N+1],maxlen;
  char A[N+1],B[N+1];
  scanf("%d",&q);
  if(q>150) exit(2);
  for(k=0;k<q;k++){
    scanf("%s %s",&A[1],&B[1]);
    A[0]=B[0]=' ';
    lenA=strlen(&A[1]);
    lenB=strlen(&B[1]);
    maxlen=0;
    for(i=0;i<=lenA;i++) count[i][0]=0;
    for(i=1;i<=lenB;i++) count[0][i]=0;
    for(i=1;i<=lenA;i++){
      for(j=1;j<=lenB;j++){
	if(A[i] == B[j]) count[i][j] = count[i-1][j-1]+1;
	else count[i][j] = max(count[i][j-1],count[i-1][j]);
	maxlen = max(maxlen,count[i][j]);
      }
    }
    printf("%d\n",maxlen);
  }
  
  return 0;
}

int max(int x,int y){
  if(x>=y)return x;
  else return y;
}