The most of players solve 6 problems, but there are only few players solve G.

haha

No if I'm not a human.

I mean yes, there's people.

Don't post irrelevant comments,especially by langugue which isn't used by most people.

Easy D,F.Hard C,E,G.

Easy F (if you have template), Hard D and E.

You're right

Disagree.I think E is hard too.

In the editorial its mentioned...

This is a widely known trick when considering Manhattan distance: we can regard that we rotated the plane by 45 degrees.

I don't know about it, and I want to know it in detail, want to solve more question related to it. Can you please share ?

Hey I tried to do a conventional dfs by considering the thing as a graph, but I got WA. I don't see what could go wrong, it passed half the test cases.

Could you help?

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;

using ld = long double;
#define int long long int
using vi = vector<int>;
using vpii = vector<pair<int,int>>;
using vvi = vector<vi>;

#define pb push_back
#define sz(x) (int)(x).size()
#define loop(i, a, b) for(int i = a; i < b; i++)
#define rloop(i, a, b) for(int i = b-1; i >= a; i--)
#define F first
#define S second
#define print(...) print_helper(__VA_ARGS__)
#define sum_(v) accumulate((v).begin(), (v).end(), 0LL)
#define min(a, b) find_min(a, b)
#define max(a, b) find_max(a, b)

#define unordered_map gp_hash_table
#define unordered_set gp_hash_table

template<typename T>
void print_helper(T t) { cout << t << "\n"; }

template<typename T, typename... Args>
void print_helper(T t, Args... args) {
    cout << t << " ";
    print_helper(args...);
}

int find_min(int a, int b) { return (a < b) ? a : b; }
int find_max(int a, int b) { return (a > b) ? a : b; }

void dfs(int source,vi &visited,vector<vector<pair<int,int>>> &nbrs,vi &result){
    visited[source]=1;
    int i=0;
    while(i<sz(nbrs[source])){
        int j=i;
        int y=nbrs[source][i].F;
        while(j<sz(nbrs[source]) && nbrs[source][j].F==y){
            int nbr=nbrs[source][j].S;
            if(!visited[nbr]){
                result.pb(nbr);
            }
            j++;
        }
        for(int k=i;k<j;k++){
            int nbr=nbrs[source][k].S;
            if(!visited[nbr]){
                dfs(nbr,visited,nbrs,result);
            }
        }
        i=j;
    }
}


signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif

    int n;
    cin >> n;
    vector<vector<pair<int, int>>> edges(n+1);
    vi result;
    loop(i,0,n){
        int x, y;
        cin >> x >> y;
        edges[x].pb({y, i+1});
    }
    loop(i,0,n+1)
    sort(edges[i].begin(),edges[i].end());
    vi visited(n+1, 0);
    dfs(0, visited, edges, result);
    for(int x : result) cout << x << " ";
    cout << endl;
    return 0;
}

Trie

weird... but true... I use that basic recursive segtree and long long for most part, I stress test too with the accepted solution of participant to find bug, but couldn't find anything!!!!!! that make mine gave wa(bug in code)... 😭 so I will have to forever live not knowing why and have to always paranoid that 💀 my segment tree implementation was maybe wrong & actually bug all along and could fuck me up at some other random time like this 🥀🥀, maybe.. maybe not, I will never know why its bug..., this is truly a curse, I think I'm gonna have PTSD now every time I use it...

Still recall big shot.

yeah it was given that xi < i

I checked the submission records for problem G. Some of them contained heavily commented code that aligns with my approach, but the implementation, as I expected, was extremely complex. I don't think I could have completed them during the contest.

However, those participants using AI have confirmed that my approach is correct.

Translate by deepseek.

There's no need for any set. Four segment trees are enough.

Build the 4 segment trees (s1, s2, s3, s4) such that

s1[i] = +x+y

s2[i] = +x-y

s3[i] = -x+y

s4[i] = -x-y

And then you can compute the queries by querying minimum on [L, R] over all segment trees.

Oh, I got it

What is "human view"?