Submission #1301507

#	Time	Username	Problem	Language	Result	Execution time	Memory
1301507		ProtonDecay314	Aliens (IOI16_aliens)	C++17	0 / 100	1 ms	344 KiB

aliens

#include "aliens.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef vector<vpll> vvpll;
typedef vector<bool> vb;

#define INF(dt) numeric_limits<dt>::max()
#define fi first
#define se second

struct Line {
    int m;

    ll b;

    int b2;

    inline pair<ll, int> evaluate(ll x) const {
        return {m * x + b, b2};
    }
};

/*
Suppose the penalty for using another picture is +y
*/
pll solve_lambda_brute(const vpll& pts, ll m, ll y) {
    ll np = pts.size();
    
    // note: stores NEGATIVE area, num pictures
    // then, the DP just gets the maximum state
    vector<pair<ll, int>> dp(np, {0ll, 0});

    deque<Line> ls;

    for(ll i = 0ll; i < np; i++) {
        ll cl = pts[i].fi, cr = pts[i].se;

        Line nl;
        if(i == 0ll) {
            nl = {- ((int)(pts[i].fi) << 1), y + pts[i].fi * pts[i].fi, -1};
        } else {
            ll overlap = max(pts[i - 1ll].se - pts[i].fi + 1ll, 0ll);
            nl = {- ((int)(pts[i].fi) << 1), dp[i - 1ll].fi - overlap * overlap + y + pts[i].fi * pts[i].fi, dp[i - 1ll].se - 1};
        }

        ls.push_back(nl);

        while((ll)(ls.size()) >= 2ll) {
            Line front = ls.front();
            ls.pop_front();

            if(front.evaluate(cr + 1ll) >= ls.front().evaluate(cr + 1ll) && front.evaluate(cr + 1ll) >= nl.evaluate(cr + 1ll)) {
                ls.push_front(front);
                break;
            }
        }

        pair<ll, int> opt = ls.front().evaluate(cr + 1ll);

        dp[i] = {opt.fi + (cr + 1ll) * (cr + 1ll), opt.se};

        // cerr << y << " " << i << " " << dp[i].fi << " " << -dp[i].se << endl;
    }

    return dp[np - 1ll];
}

ll take_photos(int i_n, int i_m, int i_k, vi r, vi c) {
    ll n = i_n, m = i_m, k = i_k;

    // first, we preprocess. remove all redundant points

    vpll pts;

    for(ll i = 0ll; i < n; i++) {
        if(r[i] > c[i]) swap(r[i], c[i]);

        pts.push_back({r[i], c[i]});
    }

    sort(pts.begin(), pts.end(), [](pll p1, pll p2){return p1.fi < p2.fi || (p1.fi == p2.fi && p1.se > p2.se);});

    ll max_r = -1ll;

    vpll filt_pts;

    for(ll i = 0ll; i < n; i++) {
        ll cl = pts[i].fi, cr = pts[i].se;

        if(cr > max_r) {
            filt_pts.push_back(pts[i]);
            // cerr << cl << " " << cr << endl;
            max_r = cr;
        }
    }

    // Performing lagrangian relaxation
    ll lo = 0ll, hi = 1e11;

    while(hi - lo > 1ll) {
        ll mid = (lo + hi) >> 1ll;

        pll cpair = solve_lambda_brute(filt_pts, m, mid);

        ll c_k = -cpair.se;

        if(c_k >= k) lo = mid;
        else hi = mid;
    }

    return solve_lambda_brute(filt_pts, m, lo).fi - k * lo;
}

Compilation message (stderr)

aliens.h:1:9: warning: #pragma once in main file
    1 | #pragma once
      |         ^~~~
aliens_c.h:1:9: warning: #pragma once in main file
    1 | #pragma once
      |         ^~~~

• Subtask 1: 0 / 4
• Subtask 2: 0 / 12
• Subtask 3: 0 / 9
• Subtask 4: 0 / 16
• Subtask 5: 0 / 19
• Subtask 6: 0 / 40