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++14 (judge verdict: Accepted)

#include <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <iomanip>
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <math.h>
#include <cmath>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
const int INF = 1001001000;
const int MOD = 1000000007;

long long gcd(long long a, long long b) {
    if (b == 0) return a;
    else return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
    return a * b / gcd(a, b);
}

using Graph = vector<vector<int>>;
//再帰とかグラフを使うときは予めNの最大値のグローバル配列を取っておく
int N;
vector<int> E[200001];
bool shipment[15][15];
vector<bool> seen;
typedef pair<int, int> P;
int D[101];
//頂点状態を表すcolor[]において、0は未訪問、1は訪問中、2は完了を表す。
int color[101];
int find_time = 0;
int dp[1001][1001];

void solve_LCS(string X,string Y) {
    int S = X.size();
    int T = Y.size();
    for (int i = 0; i < S; i++) {
        for (int j = 0; j < T; j++) {
            //同じ文字だったら、max(今まで+1,今までの最大)をする
            if (X[i] == Y[j]) dp[i + 1][j + 1] = max({ dp[i][j] + 1,dp[i + 1][j],dp[i][j + 1] });
            else dp[i + 1][j + 1] = max(dp[i + 1][j], dp[i][j + 1]);
        }
    }

    cout << dp[S][T] << endl;
}

int main() {
    int Q; cin >> Q;
    vector<string> Xs(Q);
    vector<string> Ys(Q);
    for (int i = 0; i < Q; i++) {
        cin >> Xs.at(i) >> Ys.at(i);
    }
    for (int i = 0; i < Q; i++) {
        solve_LCS(Xs.at(i),Ys.at(i));
    }
}