Submission #956048

#	Time	Username	Problem	Language	Result	Execution time	Memory
956048		shezitt	Papričice (COCI20_papricice)	C++17	110 / 110	161 ms	23720 KiB

papricice

#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <cassert>

using namespace std;

using ll = long long;

#define int ll
#define fore(i, a, b) for(int i=a; i<b; ++i)
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define pb(x) push_back(x)
#define vi vector<int>

const int N = 2e5+5;

vi adj[N];

int n;

int tam[N];

void dfs(int x, int prev=-1){
    tam[x] = 1;
    for(int child : adj[x]){
        if(child == prev) continue;
        dfs(child, x);
        tam[x] += tam[child];
    }
}

vi closest(int x, set<int> &A){
    vi res;
    auto it = A.lower_bound(x);
    if(it != A.end()){
        res.pb(*it);
    }
    if(it != A.begin()){
        it--;
        res.pb(*it);
    }
    return res;
}

int balance(int x, int y, int z){
    return max({x, y, z}) - min({x, y, z});
}

int f(int x, set<int> &A, set<int> &B, int pre=-1){

    // A -> Es ancestro
    // B -> No es ancestro
    int res = 1e9;

    // primer caso
    for(int p : closest((n + tam[x]) / 2, A)){
        res = min(res, balance(tam[x], p-tam[x], n-p));
    }

    // segundo caso
    for(int p : closest((n - tam[x]) / 2, B)){
        res = min(res, balance(tam[x], p, n-tam[x]-p));
    }

    A.insert(tam[x]);
    for(int child : adj[x]){
        if(child == pre) continue;
        res = min(res, f(child, A, B, x));
    }

    A.erase(tam[x]);
    B.insert(tam[x]);

    return res;

}

signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> n;

    fore(i, 0, n-1){
        int u, v;
        cin >> u >> v;
        adj[u].pb(v);
        adj[v].pb(u);
    }

    dfs(1);

    set<int> A, B;

    int ans = f(1, A, B, -1);

    cout << ans;

}

• Subtask 1: 15 / 15
• Subtask 2: 35 / 35
• Subtask 3: 60 / 60