Yay, hope to solve E and F.

Good Luck!Hope F!

Bad luck and low rating!

Good luck!- Hope to solve F and G.

for problem E , what's the part of my solution that giving a TLE ? is it constructing the graph ??

I don't understand. Why does my code WA on 4 testcases?

Can anyone help me why I am getting WA on 2 testcases for this solution of F?

what's the idea behind F?

Hello! I'm just a mediocre problem setter who is wondering how to give many many many D&C NTT problems which has almost the same solution quickly. May Atcoder Beginner Contest help?

UPD: I don't know whether ironies are just unwelcomed in codeforces or it's not the right place to discuss a problem which has only 143 solves with many newbies? Please, tell me the reason if you decide to downvote me @_@.

UPD 2: Maybe next time I'll try "DCNTT in ABC G author's solution, are you retarded?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????".

loved F, thank you for contest!

Solved D with a segtree

All I can see now are segment trees

"Yes" != ("YES" || "yes").

LOL, still case sensitive on output strings.

Randomized solution for F gave me some hope for a moment

For E, I forgot to handle this disconnected case:

4 2
2 7
1 2
2 8
3 4

My submission still gets AC, even though I didn't output -1 for the above case.

Could someone help me debug E? I've been absolutely confused for the second part of the contest and still can't understand where the WA is coming from: https://atcoder.jp/contests/abc352/submissions/53123170

Good luck !!!

Speedforces and Speedcoder :)

E Submission

What's wrong here?

#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
const int N = 2e6 + 10;

struct Dsu{
    vector<ll> parent, rank;
    ll sz;
    Dsu(ll n){
        sz = n;
        parent.resize(n+5); rank.resize(n+5, 1); iota(begin(parent), end(parent),0);
    }
    vector<ll> getparent() {return parent;}

    vector<ll> getrank() {return rank;}

    ll find(ll x){
        if(parent[x] == x) return x;
        return parent[x] = find(parent[x]);
    }

    void Union(ll x, ll y){
        x = find(x); y = find(y);
        if(x != y){
            if(rank[x] < rank[y])
                swap(x,y);
            rank[x] += rank[y];
            parent[y] = x;
        }
    }
    
    bool isConnected(ll x, ll y) {return find(x) == find(y);}

    ll Components(){
        ll ans = 0;
        for(ll i=1;i<=sz;i++) if(find(i) == i) ans++;
        return ans;
    }
};


int main() {
	int t;
    // cin >> t;
    t = 1;
    while(t--){
        int n, m;
        cin >> n >> m;
        map<ll, vector<ll>> M;
        for(ll i = 0; i < m; i++){
            ll k, c;
            cin >> k >> c;
            for(ll j = 0; j < k; j++){
                ll x;
                cin >> x;
                M[c].push_back(x);
            }
        }
        Dsu t = Dsu(N);
        ll sum = 0;
        for(auto [x, y]: M){
            vector<ll> cur = y;
            for(int i = 1; i < cur.size(); i++){
                if(!t.isConnected(cur[i - 1], cur[i])){
                    sum += x;
                    t.Union(cur[i - 1], cur[i]);
                }
            }
        }
        cout << (t.Components() > 1 ? -1 : sum) << '\n';
    }
}

When you join the contest late by $$$5$$$ minutes and get $$$F$$$ accepted after the contest end by $$$4$$$ minutes:

I would like to share my solution to problem F(taking me about 90 minutes to get accepted), as follows:

1. Divide all the nodes(people) into several connected components, and suppose that there are cnt of them

2. For each component, compute all the feasible rankings, and denote them based on bitmasks

3. Use dp[i][j] to denote that, for the previous i components, whether we can achieve the state of j or not, and also use set last[i][j] to store the previous states that can reach j. Here, the state means the bitmask of rankings

4. The final state should be (1<<n)-1, and we start from last[cnt][(1<<n)-1], and check whether the current state is unique or not. If it is unique, then we have unique rankings for the nodes within this component

can anyone figure out where i am going wrong in E. https://atcoder.jp/contests/abc352/submissions/53149749

Can anyone give a hint on how to optimize problem E?

I did understand the part where we could build a graph from all the given weights (using DSU) and then apply Kruskal to find the MST, but I saw the limits on K and understood this solution would result in TLE.

Can someone help me with problem E? Why my code is giving wrong answer on some test cases, even though I have also used the same logic as most others. Submission

to solve $$$E$$$ do I need to learn any algorithm?

this contest was totally fixed and made me bored. abc352 is one of the boring contests ever. abc352 fixing and boring. abc353 is genuine and much interesting with sigma problems.