因为a需要作为一个字串，b作为一个子序列

所以b里的字母可以分散在字符串中

我们需要找到b里的字母在a中分散重合的最大数量，假设为res
(求法：枚举，枚举重合第一个字母的起点)

则最短的字符串长度就是a.size() + b.size() - res

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include "bits/stdc++.h"
using namespace std;
#define ios ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define fios ofstream("test.txt");cout.rdbuf(out.rdbuf())
#define endl "\n"
#define INF 0x3f3f3f3f
#define MINF 2147483647
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) (x & (-x))
typedef unsigned long long ULL;
typedef long long LL;
typedef pair<int, int> PII;
#define x first
#define y second
const int N = 1010;


int f[N][N];

int main()
{
    int T;
    cin >> T;
    while(T--)
    {
        memset(f, 0, sizeof f);
        string a, b;
        cin >> a >> b;

        int n = a.size(), m = b.size();
        a = " " + a;
        b = " " + b;

        int res = 0;
        for(int i = 1; i <= m; i++)
        {
            int cnt = 0;
            for(int j = 1; j <= n; j++)
            {
                if(b[i + cnt] == a[j])
                {
                    cnt++;
                    res = max(res, cnt);
                }
            }
        }

        cout << n + m  - res << endl;
    }

    return 0;
}