Problem link

I implemented a fairly straightforward soln for this problem that is very similar to the editorial in essense. It uses bitsets and the main logic goes as follows:

As all ancestors for a node (lets say node a) are known when the node is first created, we can determine the nodes "b" for which a and b could form the middle part of a diamond. Whenever a new class is created, I first check if any pair of ancestors for the new class form such a couple (I denote below as "hate[][]"). If so this would create a diamond and we can terminate our search.

Otherwise I will proceed to calculate new such pairs having the new node as "a". I believe the impl is not too incomprehensible but I will be happy to answer questions. This code gets 27 points so I dont believe anything is wrong with the parsing. here is my submission code:

#include <bits/stdc++.h>
#define int long long
#define pii pair<int,int>
#define vi vector<int>
#define ff first
#define ss second
#define all(x) x.begin(),x.end()
#define sp << " " << 
using namespace std;

const int N =1e5+10,MOD = 1e9+7,inf = 2e9;

void solve() {
    int n;
    cin >> n;
    int nd = 0;
    bitset<1001> go[1001],come[1001],hate[1001],share;
    string trash;
    getline(cin,trash);
    map<string,int> mp;
    for (int i = 1;i<=n;i++) {
        string name,realname;
        getline(cin,name);
        string cur;
        int start = 0;
        vi pars;
        bool fl = 1;
        for (int j = 0;j<name.size();j++) {
            if (name[j] == ':') start = 1;
            if (!start && name[j] >= 'a' && name[j] <= 'z') realname+=name[j];
            if (start && name[j] >= 'a' && name[j] <= 'z') cur+=name[j];
            else {
                if (!cur.empty()) {
                    if (!mp.count(cur)) {
                        fl = 0;
                        break;
                    }
                    pars.push_back(mp[cur]);
                }
                cur.clear();
            }
        }
        if (!fl || mp[realname]) {
            cout << "greska" << endl;
            continue;
        }
        ++nd;
        come[nd].set(nd);
        go[nd].set(nd);
        for (auto it : pars) come[nd]|=come[it];
        for (int j = 1;j<nd && fl;j++) {
            if (!come[nd][j]) continue;
            for (int jj = j+1;jj<nd && fl;jj++) {
                if (come[nd][jj] && hate[j][jj]) {
                    fl = 0;
                    break;
                }
            }
        }
        if (!fl) {
            come[nd].reset();
            go[nd].reset();
            --nd;
            cout << "greska" << endl;
            continue; 
        }
        cout << "ok" << endl;
        for (int j = 1;j<nd;j++) {
            if (come[nd][j]) go[j][nd] = 1;
        }
        mp[realname] = nd;
        //cout << realname << " IS " << nd << '\n';
        share.reset();
        for (int j = 1;j<nd;j++){
            if (come[nd][j]) {
                share|=go[j];
            }
        }
        for (int j = 1;j<nd;j++) {
            if (share[j] && !go[j][nd]) {
                hate[nd][j] = hate[j][nd] = 1;
            }
        }
    }
} 

signed main() {
    ios_base::sync_with_stdio(0);cin.tie(0);
    #ifdef Dodi
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int t = 1;
    //cin >> t;
    while (t --> 0) solve();
}