Submission #615170

#	Time	Username	Problem	Language	Result	Execution time	Memory
615170		cheissmart	Holiday (IOI14_holiday)	C++14	24 / 100	75 ms	65536 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 = 1e9 + 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 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 = INF) {
    if(tr - tl == 1) {
        node* res = new node();
        res -> cnt = CNT(t) + 1;
        res -> sum = SUM(t) + 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));
}

void myassert(const bool b) {
    while(!b);
}

ll qqry(node* t, int need, int tl = 0, int tr = INF) {
    if(need == 0) return 0LL;
    myassert(CNT(t) >= need);
    if(tr - tl == 1)
        return 1LL * 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));
}

struct DS {
    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);

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

        for(int i = 0; i < n; i++) {
            while(SZ(stk) >= 2 && 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) {
        myassert(stk[0].F == 0);
        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;
}

• Subtask 1: 7 / 7
• Subtask 2: 0 / 23
• Subtask 3: 17 / 17
• Subtask 4: 0 / 53