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 MAX 1000
  
int main(){
  int i,j,m,q,x,y,c[MAX+1][MAX+1];
  char X[MAX],Y[MAX];
  
  scanf("%d",&q);
  
  for(i=0;i<strlen(X);i++)
    c[i][0] = 0;
  for(j=0;j<strlen(Y);j++)
    c[0][j] = 0;
  
  for(m=0;m<q;m++){
    scanf("%s",X);
    scanf("%s",Y);
    x = strlen(X);
    y = strlen(Y);
    for(i=0;i<x;i++)
      c[i][0] = 0;
    for(j=0;j<y;j++)
      c[0][j] = 0;
  
    for(i=1;i<=x;i++){
      for(j=1;j<=y;j++){
    if(X[i-1] == Y[j-1])c[i][j] = c[i-1][j-1] + 1;
    else if(c[i-1][j] >= c[i][j-1])c[i][j] = c[i-1][j];
    else c[i][j] = c[i][j-1];
      }
    }
    printf("%d\n",c[x][y]);
  }
  
  return 0;
}