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 LCS(char *, char *);
int max(int, int);
int main(){
  int q, i;
  char X[1000], Y[1000]; 
  int Z[150];
  scanf("%d", &q);
  for(i = 0; i < q; i++){
    scanf(" %s", X);
    scanf(" %s", Y);
    Z[i] = LCS(X, Y);
  }
  for(i = 0; i < q; i++){
    printf("%d\n", Z[i]);
  }
  return 0;
}


int LCS(char *x, char *y){
  int i, j, a, b;
  int lcs[1001][1001];
  a = strlen(x);
  b = strlen(y);
  for(i = 0; i < a; i++){
    lcs[i][0] = 0;
  }
  for(j = 0; j < b; j++){
    lcs[0][j] = 0;
  }

  for(i = 1; i <= a; i++){
    for(j = 1; j <= b; j++){
      if(x[i-1] == y[j-1]) lcs[i][j] = lcs[i-1][j-1]+1;
      else lcs[i][j] = max(lcs[i][j-1], lcs[i-1][j]);
    }
  }
  return lcs[a][b];
}

int max(int a, int b){
  if(a > b) return a;
  else return b;
}