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 <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define len_max 1000


string X;
string Y;
int dp[len_max+1][len_max+1];

int lcs(){
    int m = X.length();
    int n = Y.length();

    X = ' ' + X;//ずらす
    Y = ' ' + Y;
    

    for(int i = 0; i <= n; i++){
        dp[i][0] = 0;
    }

    for(int i = 0; i <= m; i++){
        dp[0][i] = 0;
    }

    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(Y[i] == X[j]){
                dp[i][j] = dp[i-1][j-1] + 1;
            }
            else{
                dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
            }
        }
    }
    return dp[n][m];

}


int main(){
    int q;
    cin >> q;

    int ans[q];

    for(int i = 0; i < q; i++){
        cin >> X >> Y;
        //printf("%d\n", lcs());
        ans[i] = lcs();
    }

    for(int i = 0; i < q; i++){
        printf("%d\n", ans[i]);
    }

    return 0;
}