Problem:
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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>

char s1[2000];
char s2[2000];
//int answer[100][100];

int LCS(int,int);

int main(){

  int sumstring;
  int i,j,k;
  scanf("%d",&sumstring);

  for(i = 0; i < sumstring; i++){

    
    scanf("%s",s1);
    scanf("%s",s2);
    // strcat(empty,s1);
    // strcat(empty,s2);
    // sprintf(s1,"%c%s",empty,s1);
    //sprintf(s2,"%c%s",empty,s2);
    //    printf("%c\n",s1[0]);
    // printf("%s\n",s2);

    // printf("%c*%c:%c\n",s1[0],s1[1],s1[2]);
    // printf("%c*%c:%c\n",s2[0],s2[1],s2[2]);

    printf("%d\n",LCS(strlen(s1),strlen(s2)));
  }

  /*  for(i = 0; i < sumstring; i++){

        printf("%d\n",ans[i]);

	}*/

  return 0;
}

int LCS(int length1, int length2){

  int i,j,x;
  int answer[1001][1001];

  x = 0;

  for(i = 0; i < length1; i++){

    answer[i][0] = 0;

  }

  for(i = 0; i < length2; i++){

    answer[0][i] = 0;

  }


  for(i = 0; i < length1; i++){

    for(j = 0; j < length2; j++){

      if(s1[i]==s2[j] && s1[i]!=' ' && s2[i] != ' '){

	if(i == 0 || j == 0){

	  answer[i][j] = 1;
     
	}
	else{
	  //	printf("%c:::%c\n",s1[0],s2[0]);
	answer[i][j] = answer[i-1][j-1]+1  ;
	
	
	}
      }


      else{


	if(i == 0 && j != 0){
	  answer[i][j] = answer[0][j-1];
	}
	
	else if(i !=0 && j == 0){

	  answer[i][j] = answer[i-1][0];

	}

	else if(i == 0 && j == 0){

	  answer[0][0] = 0;

	}

	else if(i != 0 && answer[i-1][j] > answer[i][j-1]){

	  //  printf("**%d::\n",answer[i-1][j]);
	  answer[i][j] = answer[i-1][j];
	  
	}

	else if(answer[i-1][j] <= answer[i][j-1]){

	  answer[i][j] = answer[i][j-1];

	}
	
      }
    }
  }

  return answer[length1-1][length2-1];


  /* for(i = 0; i < length1; i++){

    for(j = 0; j < length2; j++){

      //printf("%d\n",length2);
      //      printf("%d:%d:%d\n",i,j,answer[i][j]);
      if(x < answer[i][j]){

	x = answer[i][j];

      }

    }

  }

  return x;

  */
}