Submission #329266

#	Time	Username	Problem	Language	Result	Execution time	Memory
329266		534351	Papričice (COCI20_papricice)	C++17	110 / 110	293 ms	33388 KiB

papricice

#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>

using namespace std;
// using namespace __gnu_pbds;

// template<class T>
// using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

const int MAXN = 200013;

int N;
vi edge[MAXN];
int parent[MAXN], subtree[MAXN];
set<int> sizes[MAXN];
int ans;
vi nums;

void dfs(int u, int p)
{
    subtree[u] = 1;
    for (int v : edge[u])
    {
        if (v == p) continue;
        parent[v] = u;
        dfs(v, u);
        subtree[u] += subtree[v];
    }
}

//case 0: they're in same subtree

int calc(int a, int b)
{
    int c = N - a - b;
    // cerr << "calc " << a << ' ' << b << endl;
    return max(a, max(b, c)) - min(a, min(b, c));
}
void solve(set<int> &a, set<int> &b)
{
    for (int x : a)
    {
        auto it = b.LB((N - x) / 2);
        if (it != b.end())
        {
            ckmin(ans, calc(x, *it));
        }
        if (it != b.begin())
        {
            ckmin(ans, calc(x, *prev(it)));
        }
    }
    return;
}
//case 1: they're in two separate subtrees
void solve1(int u)
{
    for (int v : edge[u])
    {
        if (v == parent[u]) continue;
        solve1(v);
    }
    for (int v : edge[u])
    {
        if (v == parent[u]) continue;
        if (SZ(sizes[v]) > SZ(sizes[u])) swap(sizes[u], sizes[v]);
        solve(sizes[u], sizes[v]);
        for (int x : sizes[v]) sizes[u].insert(x);
    }
    sizes[u].insert(subtree[u]);
    while(!sizes[u].empty() && *sizes[u].begin() * 3 + 2 * ans <= N) sizes[u].erase(sizes[u].begin());
    while(!sizes[u].empty() && *prev(sizes[u].end()) * 3 - 2 * ans >= N) sizes[u].erase(prev(sizes[u].end()));
}
void solve2(int u)
{
    nums.PB(subtree[u]);
    for (int v : edge[u])
    {
        if (v == parent[u]) continue;
        solve2(v);
    }
    nums.pop_back();
    int sz = (N + subtree[u]) / 2;
    auto it = LB(ALL(nums), sz, greater<int>());
    if (it != nums.end())
    {
        ckmin(ans, calc(subtree[u], *it - subtree[u]));
    }
    if (it != nums.begin())
    {
        ckmin(ans, calc(subtree[u], *prev(it) - subtree[u]));
    }
    return;
    //now solve: one cut is subtree[u], the other cut is smth before
}

int32_t main()
{
    cout << fixed << setprecision(12);
    cerr << fixed << setprecision(4);
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> N;
    FOR(i, 0, N - 1)
    {
        int u, v;
        cin >> u >> v; u--; v--;
        edge[u].PB(v);
        edge[v].PB(u);
    }
    parent[0] = N;
    dfs(0, N);
    ans = N;
    solve2(0);
    solve1(0);
    cout << ans << '\n';
    return 0;
}

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