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 SIZE 1000

int max(int a,int b){
  return (a>b ? a:b);
}

int main(){
  int n,i,j,cnt=0;
  char x[SIZE],y[SIZE];
  scanf("%d",&n);
  
  while(cnt != n){
    char tmp1[SIZE],tmp2[SIZE];
    scanf("%s %s",tmp1,tmp2);
    x[0]=y[0]=' ';
    strcpy(x+1,tmp1);
    strcpy(y+1,tmp2);
    int dp[SIZE+100][SIZE+100];
    int xlen = strlen(x);
    int ylen = strlen(y);
    int ans = 0;

    for(i=0; i<=xlen; i++) dp[i][0] = 0;
    for(i=0; i<=ylen; i++) dp[0][i] = 0;
    
    for(i=1; i<=xlen; i++){
      for(j=1; j<=ylen; j++){
        if(x[i] == y[j]){
          dp[i][j] = dp[i-1][j-1]+1;
                }
        else{
          dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
        }
        ans = max(ans,dp[i][j]);
      }
    }
    
    ans--;
        
    printf("%d\n",ans);
    
    cnt++;
  }
  
  return 0;
}