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 <stdlib.h>
#include <stdio.h>
#define MAX_ARR_SIZE 1024

static int count_lcs(char *x, char *y) {
  int i, j, m, n, max, mn[MAX_ARR_SIZE][MAX_ARR_SIZE];

  for (i = m = 0; x[i] != '\0'; ++i) ++m;
  for (j = n = 0; y[j] != '\0'; ++j) ++n;
  for (i = 0; i < m; ++i) mn[i][0] = 0;
  for (j = 0; j < n; ++j) mn[0][j] = 0;
  max = 0;
  for (i = 1; i < m + 1; ++i) {
    for (j = 1; j < n + 1; ++j) {
      if (x[i - 1] == y[j - 1]) {
        mn[i][j] = mn[i - 1][j - 1] + 1;
      } else {
        if (mn[i][j - 1] > mn[i - 1][j]) {
          mn[i][j] = mn[i][j - 1];
        } else {
          mn[i][j] = mn[i - 1][j];
        }
      }
      max = mn[i][j] > max ? mn[i][j] : max;
    }
  }
  return max;
}

int main(int argc, char **argv) {
  int q, i;
  char x[MAX_ARR_SIZE], y[MAX_ARR_SIZE];

  scanf("%d", &q);
  for (i = 0; i < q; ++i) {
    scanf("%s", x);
    scanf("%s", y);
    printf("%d\n", count_lcs(x, y));
  }
  exit(0);
}