Submission #546780

#	Time	Username	Problem	Language	Result	Execution time	Memory
546780		gimabd30	Fireworks (APIO16_fireworks)	C++17	100 / 100	180 ms	39580 KiB

fireworks

//#define _GLIBCXX_DEBUG

//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <bits/stdc++.h>

using namespace std;

#include <ext/pb_ds/assoc_container.hpp>

using namespace __gnu_pbds;

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

template<typename T>
using normal_queue = priority_queue <T, vector<T>, greater<>>;

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

#define ll long long
#define trace(x) cout << #x << ": " << (x) << endl;
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define uniq(x) x.resize(unique(all(x)) - begin(x))
#define ld long double
#define sz(s) ((int) size(s))
#define pii pair<int, int>
#define mp(x, y) make_pair(x, y)
#define int128 __int128
#define pb push_back
#define eb emplace_back


template<typename T>
bool ckmin(T &x, T y) {
    if (x > y) {
        x = y;
        return true;
    }
    return false;
}

template<typename T>
bool ckmax(T &x, T y) {
    if (x < y) {
        x = y;
        return true;
    }
    return false;
}

int bit(int x, int b) {
    return (x >> b) & 1;
}


int rand(int l, int r) { return (int) ((ll) rnd() % (r - l + 1)) + l; }

//soryan za musor sverhu

const ll infL = 3e18;
const int infI = 1e9 + 7;
const int N = 300001;
const ll mod = 1e9 + 7;
const ld eps = 1e-9;

priority_queue<ll> pq[N];
int p[N], c[N];
ll k[N], b[N];
//это уравнение прямой при больших X

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    //https://codeforces.com/blog/entry/47821
    int n, m;
    cin >> n >> m;
    auto mrg = [](int i) {
        if (sz(pq[i]) > sz(pq[p[i]])) swap(pq[i], pq[p[i]]);
        k[p[i]] += k[i];
        b[p[i]] += b[i];
        while (!pq[i].empty()) {
            auto v = pq[i].top();
            pq[i].pop();
            pq[p[i]].push(v);
        }
    };
    for (int i = 1; i < n + m; ++i) {
        cin >> p[i] >> c[i];
        --p[i];
    }
    for (int i = n; i < n + m; ++i) {
        k[i] = 1;
        b[i] = -c[i];
        pq[i].push(c[i]);
        pq[i].push(c[i]);
        mrg(i);
    }
    for (int i = n - 1; i > 0; --i) {
        //we don't need slopes with k > 1 because we just can increase c[i[ by one all the time
        while (k[i] > 1) {
            --k[i];
            b[i] += pq[i].top(); // y = kx+b = (a-1)x + (b + x) at slope changing point
            pq[i].pop();
        }
        auto R = pq[i].top();
        pq[i].pop();
        auto L = pq[i].top();
        pq[i].pop();
        L += c[i];
        R += c[i];
        //moving slopes with K = 0, 1
        b[i] -= c[i];
        //we shifted to the right by c[i] and k = 1 so b[i] decreases by c[i]
        pq[i].push(L);
        pq[i].push(R);
        mrg(i);
    }
    while (k[0] > 0) {
        --k[0];
        b[0] += pq[0].top();// y = kx+b = (a-1)x + (b + x) at slope changing point
        pq[0].pop();
    }
    cout << b[0];
    return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 19 / 19
• Subtask 3: 29 / 29
• Subtask 4: 45 / 45