제출 #1123734

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1123734		math_rabbit_1028	나일강 (IOI24_nile)	C++20	100 / 100	279 ms	24324 KiB

nile

#include "nile.h"
#include <queue>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;

int N;
struct minseg {
    ll tree[404040];
    void init(int v, int st, int ed) {
        if (st == ed) {
            tree[v] = 1e18;
            return;
        }
        int mid = (st+ed)/2;
        init(2*v, st, mid);
        init(2*v+1, mid+1, ed);
        tree[v] = min(tree[2*v], tree[2*v+1]);
    }
    void update(int v, int st, int ed, int idx, ll val) {
        if (st > idx || ed < idx) return;
        if (st == idx && ed == idx) {
            tree[v] = min(tree[v], val);
            return;
        }
        int mid = (st+ed)/2;
        update(2*v, st, mid, idx, val);
        update(2*v+1, mid+1, ed, idx, val);
        tree[v] = min(tree[2*v], tree[2*v+1]);
    }
    ll get(int v, int st, int ed, int lt, int rt) {
        if (lt > ed || rt < st) return 1e18;
        if (lt <= st && ed <= rt) return tree[v];
        int mid = (st+ed)/2;
        return min(get(2*v, st, mid, lt, rt), get(2*v+1, mid+1, ed, lt, rt));
    }
} o1, o2, e1, e2;

vector<pll> C;
ll block(int st, int ed) { // [st, ed]
    if ((ed-st)%2 == 1) return 0;
    if (st%2) return min(o1.get(1, 0, N-1, st, ed), e2.get(1, 0, N-1, st, ed));
    else return min(e1.get(1, 0, N-1, st, ed), o2.get(1, 0, N-1, st, ed));
}

vector<ll> calculate_costs(vector<int> W, vector<int> A, vector<int> B, vector<int> E) {
    int Q = (int)E.size();
    N = (int)W.size();

    ll sum = 0;
    for (int i = 0; i < N; i++) sum += B[i];
    vector<ll> R(Q, sum);
    for (int i = 0; i < N; i++) C.push_back({W[i], A[i]-B[i]});
    sort(C.begin(), C.end());

    o1.init(1, 0, N-1);
    e1.init(1, 0, N-1); 
    for (int i = 0; i < N; i++) {
        if (i%2) o1.update(1, 0, N-1, i, C[i].second);
        else e1.update(1, 0, N-1, i, C[i].second);
    }
    o2.init(1, 0, N-1);
    e2.init(1, 0, N-1);

    priority_queue<pll, vector<pll>, greater<pll>> two, one;
    for (int i = 1; i < N-1; i++) {
        two.push({C[i+1].first - C[i-1].first, i});
    }
    for (int i = 1; i < N; i++) {
        one.push({C[i].first - C[i-1].first, i});
    }
    vector<ll> qry;
    map<ll, ll> ans;
    for (int i = 0; i < Q; i++) qry.push_back(E[i]);
    sort(qry.begin(), qry.end());
    qry.erase(unique(qry.begin(), qry.end()), qry.end());

    map<pll, ll> blocks;
    sum = 0;
    for (int i = 0; i < N; i++) {
        blocks[{i, i}] = C[i].second;
        sum += C[i].second;
    }
    for (ll D : qry) {
        while (!two.empty()) {
            if (two.top().first > D) break;
            int i = two.top().second;
            if (i%2) o2.update(1, 0, N-1, i, C[i].second);
            else e2.update(1, 0, N-1, i, C[i].second);
            auto iter = blocks.upper_bound({i, 0}); iter--;
            sum -= iter->second;
            iter->second = block(iter->first.first, iter->first.second);
            sum += iter->second;
            two.pop();
        }

        while (!one.empty()) {
            if (one.top().first > D) break;
            int i = one.top().second;
            auto now = blocks.lower_bound({i, 0});
            auto prev = now; prev--;
            sum -= now->second;
            sum -= prev->second;
            pll temp = {prev->first.first, now->first.second};
            blocks[temp] = block(prev->first.first, now->first.second);
            blocks.erase(prev);
            blocks.erase(now);
            sum += blocks[temp];
            one.pop();
        }

        ans[D] = R[0] + sum;
    }

    for (int t = 0; t < Q; t++) {
        R[t] = ans[E[t]];
    }
    return R;
}

• Subtask 0: 0 / 0
• Subtask 1: 6 / 6
• Subtask 2: 13 / 13
• Subtask 3: 17 / 17
• Subtask 4: 11 / 11
• Subtask 5: 20 / 20
• Subtask 6: 15 / 15
• Subtask 7: 18 / 18