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>
#define N 1000
#define MAX(a, b) (((a) > (b)) ? (a): (b))
int lcs(char*, char*);
int main(void)
{
  int i, n;
  char s1[N+1], s2[N+1];
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%*c%s%*c%s", s1, s2);
    printf("%d\n", lcs(s1, s2));
  }
}
int lcs(char *s1, char *s2)
{
  int i, j, maxl = 0, c[N+1][N+1], m = strlen(s1), n = strlen(s2);
  for (i = 1; i <= m; i++) c[i][0] = 0;
  for (j = 1; j <= n; j++) c[0][j] = 0;
  for (i = 1; i <= m; i++) {
    for (j = 1; j <= n; j++) {
      if (s1[i-1] == s2[j-1]) {
        c[i][j] = c[i-1][j-1] + 1;
      } else {
        c[i][j] = MAX(c[i-1][j], c[i][j-1]);
      }
      maxl = MAX(maxl, c[i][j]);
    }
  }
  return c[m][n];
}