//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h> // By Lucky Ox
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define int long long
#define double long double
#define endl "\n"
#define pii pair<int, int>
#define tans cout << "答案："
using namespace std;
//using i128 = __int128;
const int mod = 1000000000000002493, INF = 0x3f3f3f3f3f3f3f3f;
int lcm(int x, int y) { return x * y / __gcd(x, y); }
int q_pow(int a, int k) { int res = 1; while (k) { if (k & 1) res = res * a % mod; k >>= 1; a = a * a % mod; } return res; }
int len_10(int x) { int len = 0; while(x) { x /= 10; len ++ ; } return len;}
int len_2(int x) { int len = 0; while(x) { x /= 2; len ++ ; } return len;}
// n / m 上取整 = (n + m - 1) / m

const int N = 20;
int n;
pair<string, string> a[N];
int p[N], cnt[N];
int find(int x) {
    if(x != p[x]) p[x] = find(p[x]);
    return p[x];
}
void solve() {
    cin >> n;
    for(int i = 1; i <= n; i ++ ) {
        p[i] = i, cnt[i] = 1;
        cin >> a[i].first >> a[i].second;
    }

    for(int i = 1; i <= n; i ++ ) {
        for(int j = i + 1; j <= n; j ++ ) {
            int fa = find(i), fb = find(j);
            if(fa != fb && (a[i].first == a[j].first || a[i].second == a[j].second)) {
                cnt[fa] += cnt[fb];
                p[fb] = fa;
            }
        }
    }
    int maxn = 0;
    for(int i = 1; i <= n; i ++ ) maxn = max(maxn, cnt[i]);
    cout << n - maxn << endl;
}

void init() {

}

signed main() {
    IOS;
    cout << fixed << setprecision(10);
    init();
    int T = 1;
    cin >> T;
    while (T -- ) solve();
    return 0;
}

代码如上 思路就是 如果两个音乐能拼接起来 那么就归到一个集合里面 那么我们最多就只能留下一个集合 那么肯定留最大的集合 能删的最小值 应该是 n - maxn（maxn表示最大的集合内元素的数量）