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 JAVA (judge verdict: Accepted)

import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int setnums = scan.nextInt();

		for(int i=0; i<setnums; i++) {
			String str1 = scan.next();
			String str2 = scan.next();
			String longer = "";
			String smaller = "'";
			if(str1.length() > str2.length()) {
				longer = str1;
				smaller = str2;
			}else {
				smaller = str1;
				longer = str2;
			}
			longestSeq(longer, smaller);
		}
		
	}
	
	static void longestSeq(String large, String small) {
		int[][] map = new int[large.length()+1][small.length()+1];
		//System.out.println("mp: " + map[1][1]);
		int seqlength = 0;
		for(int i=0; i<=large.length(); i++){ 
	      for (int j=0; j<=small.length(); j++) { 
	        if (i == 0 || j == 0) {
	            map[i][j] = 0; 
	        } else {
	        	char largeChar = large.charAt(i-1);
	        	char smallChar = small.charAt(j-1);
	        	if (largeChar == smallChar) {
	        		map[i][j] = map[i-1][j-1] + 1; 
	        	} else {
	        		int largeVal = map[i-1][j];
	        		int smallVal = map[i][j-1];
	        		if(largeVal>smallVal){
	        			map[i][j] = map[i-1][j];
	        		} else {
	        			map[i][j] = map[i][j-1];
	        		}
	        	}
	        }
	        seqlength = map[large.length()][small.length()];
	      }
		}
        System.out.println(seqlength);

	}
}
//		for(int i=0; i<small.length(); i++) {
//		for(int j=0; j<large.length(); j++) {
//			
//		}
//	}
//	
			
//	for(int j=0; j<small.length(); j++) {
//		t1 = j;
//		boolean cont = true;
//		//for(int k = 0; k<large)
//		for(int k=0; k<large.length(); k++) {
//			t2 = k;
//		while(cont) {
//			if((small.length() > t1) && (small.charAt(t1) == large.charAt(t2))) {
//				System.out.println(small.charAt(t1) + " t1:" + t1 + " t2:" + t2);
//				current += small.charAt(t1);
//				//System.out.println("here if");
//				t1++;
//				t2++;
//			}else {
//				//System.out.println("here else");
//				cont = false;
//				t1 = 0;
//				t2 = 0;
//			}		
//			if(current.length()>longest.length()) {
//				longest = current;
//			}
//			//System.out.println("cur: " + current);
//		}
//		}
//	}
//	System.out.println("longest: " + longest);