L2-005 set similarity (set judgment)

  Ideas ： A question of comparing water , However, if you want to pass all test points, you need to consider the maximum complexity .

Code implementation ：

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
const int N = 100;

int n, m, x, k;
set<int> s[N], tmp;

signed main()
{
    cin >> n;
    for(int i = 1; i <= n; i ++){
        cin >> m;
        for(int j = 0; j < m; j ++){
            cin >> x;
            s[i].insert(x);
        }
    }
    cin >> k;
    for(int i = 0; i < k; i ++){
        int s1, s2;
        cin >> s1 >> s2;
        //  The last test point will time out 
//        tmp.clear();
//        for(auto x:s[s1]){
//            tmp.insert(x);
//        }
//        for(auto x:s[s2]){
//            tmp.insert(x);
//        }
//        int nt = tmp.size();
//        int nc = s[s1].size()+s[s2].size()-nt;
        int nc = 0, nt = s[s2].size();
        for(auto x:s[s1]){
            if(s[s2].find(x)!=s[s2].end()) nc ++;
            else nt ++;
        }
        cout << fixed << setprecision(2) << nc*100.0/nt << "%\n";
    }

    return 0;
}