제출 #615368

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
615368		cheissmart	휴가 (IOI14_holiday)	C++17	100 / 100	1542 ms	45592 KiB

holiday

#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 {
    static const int MX = 2e6;
    int l[MX], r[MX], cnt[MX], sz;
    ll sum[MX];
    vi val;

    int upd(int t, int pos, int tl, int tr) {
        int nw = sz++;
        if(tr - tl == 1) {
            cnt[nw] = cnt[t] + 1;
            sum[nw] = sum[t] + val[tl];
            return nw;
        }
        l[nw] = l[t], r[nw] = r[t];
        int tm = (tl + tr) / 2;
        if(pos < tm) l[nw] = upd(l[t], pos, tl, tm);
        else r[nw] = upd(r[t], pos, tm, tr);
        cnt[nw] = cnt[l[nw]] + cnt[r[nw]];
        sum[nw] = sum[l[nw]] + sum[r[nw]];
        return nw;
    }
 
    ll qqry(int t, int need, int tl, int tr) {
        if(need == 0) return 0LL;
        assert(cnt[t] >= need);
        if(tr - tl == 1)
            return 1LL * val[tl] * need;
        int tm = (tl + tr) / 2;
        if(cnt[r[t]] >= need)
            return qqry(r[t], need, tm, tr);
        else
            return qqry(l[t], need - cnt[r[t]], tl, tm) + sum[r[t]];
    }
 
    vi a;
    V<pi> stk;
    int c, n;
    vi 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, 0, SZ(val));
    }
    DS(vi _a, int _c) {
        memset(l, 0, sizeof l);
        memset(r, 0, sizeof r);
        memset(cnt, 0, sizeof cnt);
        memset(sum, 0, sizeof sum);

        a = _a;
        c = _c;
        n = SZ(a);
        sz = 1;
 
        val = a;
        sort(ALL(val));
        val.resize(unique(ALL(val)) - val.begin());
 
        int t = 0;
        for(int i = 0; i < n; i++) {
            int tt = lower_bound(ALL(val), a[i]) - val.begin();
            t = upd(t, tt, 0, SZ(val));
            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]);
    V<ll> aux(d);
    {
        DS dsl = DS(pl, 2);
        for(int i = 0; i + 2 <= d; i++)
            aux[i] += dsl.qry(i);
    }
    {
        DS dsr = DS(pr, 1);
        for(int i = 0; i + 2 <= d; i++)
            aux[i] += dsr.qry(d - i - 2);
    }
    ll ans = 0;
    for(int i = 0; i + 2 <= d; i++)
        ans = max(ans, aux[i]);
    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;
}

• Subtask 1: 7 / 7
• Subtask 2: 23 / 23
• Subtask 3: 17 / 17
• Subtask 4: 53 / 53