제출 #639381

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
639381		piOOE	Aliens (IOI16_aliens)	C++17	100 / 100	200 ms	7508 KiB

aliens

#include <bits/stdc++.h>
#include "aliens.h"

using namespace std;

using ll = long long;

ll sq(ll a) {
    return a * a;
}

constexpr ll inf = 3e18;

struct Line {
    ll k = 0, b = inf;
    int idx = -1;
    double last = -inf;

    ll eval(ll x) {
        return k * x + b;
    }

    double intersect(Line a) {
        return (a.b - b) / double(k - a.k);
    }
};

struct CHT {
    deque<Line> t;

    void add(Line a) {
        while (t.size() > 1 && a.intersect(t.end()[-2]) < t.back().last) {
            t.pop_back();
        }
        if (!t.empty()) {
            a.last = a.intersect(t.back());
        }
        t.push_back(a);
    }

    Line query(ll x) {
        while (t.size() > 1 && t[1].eval(x) < t[0].eval(x)) {
            t.pop_front();
        }
        return t[0];
    }
};

long long take_photos(int n, int m, int k, std::vector<int> r, std::vector<int> c) {
    for (int i = 0; i < n; ++i) {
        if (r[i] < c[i]) {
            swap(r[i], c[i]);
        }
        ++r[i];
    }
    vector<int> ord(n);
    iota(ord.begin(), ord.end(), 0);
    sort(ord.begin(), ord.end(), [&](int i, int j) {
        return r[i] != r[j] ? r[i] < r[j] : c[i] > c[j];
    });
    vector<int> stk;
    for (int i: ord) {
        while (!stk.empty() && c[stk.back()] >= c[i]) {
            stk.pop_back();
        }
        stk.push_back(i);
    }
    vector<int> nr = {-1}, nc = {-1};
    for (int i: stk) {
        nr.push_back(r[i]), nc.push_back(c[i]);
    }
    swap(r, nr), swap(c, nc);
    n = r.size();

    k = min(k, n - 1);

    ll L = 0, R = ll(m) * m + 1;
    ll ans = 0;
    while (L + 1 < R) {
        ll mid = (L + R) / 2;
        CHT t;
        vector<ll> dp(n, inf);
        dp[0] = 0;
        vector<int> cnt(n);
        for (int i = 0; i < n; ++i) {
            if (i > 0) {
                Line line = t.query(r[i]);
                dp[i] = line.eval(r[i]) + mid + sq(r[i]);
                cnt[i] = cnt[line.idx] + 1;
            }
            if (i != n - 1) {
                t.add({-2 * c[i + 1], dp[i] - sq(max(0, r[i] - c[i + 1])) + sq(c[i + 1]), i});
            }
        }
        if (cnt[n - 1] <= k) {
            ans = dp[n - 1];
            R = mid;
        } else {
            L = mid;
        }
    }

    return ans - k * R;
}

• Subtask 1: 4 / 4
• Subtask 2: 12 / 12
• Subtask 3: 9 / 9
• Subtask 4: 16 / 16
• Subtask 5: 19 / 19
• Subtask 6: 40 / 40