제출 #615213

#제출 시각아이디문제언어결과실행 시간메모리
615213cheissmart휴가 (IOI14_holiday)C++14
24 / 100
76 ms65536 KiB
#include"holiday.h"
#include <bits/stdc++.h>
#define F first
#define S second
#define V vector
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(v) int((v).size())
#define ALL(v) (v).begin(), (v).end()

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;

const int INF = 1e9 + 7, C = 1e5 + 7;


ll go(int n, int s, int d, int a[]) {
    priority_queue<int, vi, greater<int>> pq;
    ll ans = 0, sum = 0;
    int cost = 0;
    for(int i = s; i < n; i++) {
        pq.push(a[i]);
        sum += a[i];
        while(SZ(pq) + cost > d) {
            if(pq.empty()) return ans;
            sum -= pq.top();
            pq.pop();
        }
        ans = max(ans, sum);
        cost++;
    }
    return ans;
}

struct DS {

    int val[C];
    struct node {
        node *l, *r;
        int cnt;
        ll sum;
        node() {
            cnt = sum = 0;
            l = r = nullptr;
        }
        node(node *_l, node *_r) {
            l = _l, r = _r;
            cnt = (l ? l -> cnt : 0) + (r ? r -> cnt : 0);
            sum = (l ? l -> sum : 0LL) + (r ? r -> sum : 0LL);
        }
    };

    int CNT(node* t) {
        return t ? t -> cnt : 0;
    }
    ll SUM(node* t) {
        return t ? t -> sum : 0LL;
    }
    node* lc(node* t) {
        return t ? t -> l : nullptr;
    }
    node* rc(node* t) {
        return t ? t -> r : nullptr;
    }

    node* upd(node* t, int pos, int tl = 0, int tr = C) {
        if(tr - tl == 1) {
            node* res = new node();
            res -> cnt = CNT(t) + 1;
            res -> sum = SUM(t) + val[tl];
            return res;
        }
        int tm = (tl + tr) / 2;
        if(pos < tm) return new node(upd(lc(t), pos, tl, tm), rc(t));
        else return new node(lc(t), upd(rc(t), pos, tm, tr));
    }

    ll qqry(node* t, int need, int tl = 0, int tr = C) {
        if(need == 0) return 0LL;
        if(tr - tl == 1)
            return 1LL * val[tl] * need;
        int tm = (tl + tr) / 2;
        if(CNT(rc(t)) >= need)
            return qqry(rc(t), need, tm, tr);
        else
            return qqry(lc(t), need - CNT(rc(t)), tl, tm) + SUM(rc(t));
    }

    vi a;
    V<pi> stk;
    int c, n;
    V<node*> h;

    ll test(int d, int i) {
        if(i * c > d) return -1;
        d -= i * c;
        d = min(d, i + 1);
        return qqry(h[i], d);
    }
    DS(vi _a, int _c) {
        a = _a;
        c = _c;
        n = SZ(a);

        vi compress = a;
        sort(ALL(compress));
        compress.resize(unique(ALL(compress)) - compress.begin());
        for(int i = 0; i < SZ(compress); i++)
            val[i] = compress[i];

        node* t = nullptr;
        for(int i = 0; i < n; i++) {
            int tt = lower_bound(ALL(compress), a[i]) - compress.begin();
            t = upd(t, tt);
            h.PB(t);
        }

        for(int i = 0; i < n; i++) {
            while(SZ(stk) && test(stk.back().F, stk.back().S) <= test(stk.back().F, i))
                stk.pop_back();
            if(stk.empty())
                stk.EB(i * c, i);
            else {
                int lb = stk.back().F + 1, rb = INF;
                // find first x s.t. test(x, stk.back().S) <= test(x, i)
                while(lb <= rb) {
                    int mb = (lb + rb) / 2;
                    if(test(mb, stk.back().S) <= test(mb, i))
                        rb = mb - 1;
                    else
                        lb = mb + 1;
                }
                stk.EB(lb, i);
            }
        }
    }
    ll qry(int d) {
        int it = upper_bound(ALL(stk), pi(d, INF)) - stk.begin() - 1;
        return test(d, stk[it].S);
    }
};

ll solve(int n, int s, int d, int a[]) { // go left in the first step
    if(s == 0) return 0;
    vi pl, pr;
    for(int i = s - 1; i >= 0; i--) pl.PB(a[i]);
    for(int i = s; i < n; i++) pr.PB(a[i]);
    DS dsl = DS(pl, 2);
    DS dsr = DS(pr, 1);
    ll ans = 0;
    for(int i = 0; i + 2 <= d; i++) {
        ans = max(ans, dsl.qry(i) + dsr.qry(d - i - 2));
    }
    return ans;
}

ll findMaxAttraction(int n, int s, int d, int a[]) {
    ll ans = go(n, s, d, a);

    ans = max(ans, solve(n, s, d, a));

    reverse(a, a + n);
    s = n - s - 1;

    ans = max(ans, go(n, s, d, a));
    ans = max(ans, solve(n, s, d, a));

    return ans;
}
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