# 目次

• 目次
• 解答
• 解答

# 解答

タイムアウトになる解答.

// C++ 14
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <math.h>

#define ll long long
#define Int int
#define loop(x, start, end) for(Int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)
#define rep(n) for(int x = 0; x < n; x++)
#define span(a,x,y) a.begin()+x,a.begin()+y
#define span_all(a) a.begin(),a.end()
#define len(x) (x.size())
#define last(x) (*(x.end()-1))

using namespace std;

int lcs(string &s1, string &s2, int i1, int i2) {
    if (i1 >= s1.length() || i2 >= s2.length()) return 0;
    if (s1[i1] == s2[i2]) return 1 + lcs(s1, s2, i1+1, i2+1);
    return max(
        lcs(s1, s2, i1 + 1, i2),
        lcs(s1, s2, i1, i2 + 1)
    );
}

int main(void) {
    int n;
    string s1, s2;
    
    cin >> n;
    loop(i, 0, n) {
        cin >> s1 >> s2;
        cout << lcs(s1, s2, 0, 0) << endl;
    }
}

# 解答

動的計画法を使ってO(MN)まで高速化した解法.

// C++ 14
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <math.h>

#define ll long long
#define Int int
#define loop(x, start, end) for(Int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)
#define rep(n) for(int x = 0; x < n; x++)
#define span(a,x,y) a.begin()+x,a.begin()+y
#define span_all(a) a.begin(),a.end()
#define len(x) (x.size())
#define last(x) (*(x.end()-1))

using namespace std;

#define MAX 1000



int lcs(string &s1, string &s2) {
    int dp[MAX+1][MAX+1];
    int l1 = s1.length();
    int l2 = s2.length();
    int maxl = 0;
    
    s1 = ' ' + s1;
    s2 = ' ' + s2;
    
    loop(i, 0, l1+1) {
        dp[i][0] = 0;
    }
    loop(j, 0, l2+1) {
        dp[0][j] = 0;
    }
    
    loop (i, 1, l1+1) {
        loop (j, 1, l2+1) {
            if (s1[i] == s2[j]) dp[i][j] = 1 + dp[i-1][j-1];
            else dp[i][j] = max(
                dp[i-1][j],
                dp[i][j-1]
            );
            
            maxl = max(maxl, dp[i][j]);
        }
        
    }
    
    return maxl;
}

int main(void) {
    int n;
    string s1, s2;
    
    cin >> n;
    loop(i, 0, n) {
        cin >> s1 >> s2;
        cout << lcs(s1, s2) << endl;
    }
}