Submission #1057430

#	Time	Username	Problem	Language	Result	Execution time	Memory
1057430		TAhmed33	Aliens (IOI16_aliens)	C++98	60 / 100	1078 ms	1048576 KiB

aliens

#include "aliens.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
//#include "grader.cpp"
const ll inf = 1e16;
typedef long double ld;
const ld eps = 1e-9;
inline ll sq (ll x) {
    return x * x;
}
struct CHT {
    deque <pair <ll, ll>> dd;
    void init () {
    }
    ld inter (pair <ll, ll> x, pair <ll, ll> y) {
        return 1.0 * (x.second - y.second) / (y.first - x.first);   
    }
    ll query (ll x) {
        while (dd.size() >= 2) {
            if (inter(dd[dd.size() - 1], dd[dd.size() - 2]) <= x) {
                break;
            }
            dd.pop_back();
        }
        return dd.back().first * x + dd.back().second;
    }
    void insert (ll x, ll y) {
        pair <ll, ll> g = {x, y};
        while (dd.size() >= 2 && inter(g, dd[0]) > inter(dd[0], dd[1]) + eps) {
            dd.pop_front();
        }
        dd.push_front(g);
    }
};
ll take_photos (int n, int m, int k, vector <int> r, vector <int> c) {
    array <ll, 2> a[n + 2];
    for (int i = 0; i < n; i++) {
        if (r[i] < c[i]) {
            swap(r[i], c[i]);
        }
        c[i]++; r[i]++;
        a[i + 1] = {c[i], r[i]};
    }
    sort(a + 1, a + n + 1, [&] (array <ll, 2> &x, array <ll, 2> &y) {
        return x[0] == y[0] ? x[1] > y[1] : x[0] < y[0];
    });
    ll mx = 0;
    array <ll, 2> b[n + 2];
    int d = 0;
    for (int i = 1; i <= n; i++) {
        if (mx >= a[i][1]) {
            continue;
        }
        b[++d] = a[i];
        mx = a[i][1];
    }
    n = d;
    ll dp[n + 2][k + 2] = {};
    for (int i = 1; i <= n; i++) {
        a[i] = b[i];
    }
    a[n + 1] = {m + 1, m + 1};
    for (int i = 1; i <= n; i++) {
        dp[i][0] = inf;
    }
    for (int j = 1; j <= k; j++) {
        CHT cur;
        cur.init();
        for (int i = n; i >= 1; i--) {
            cur.insert(-2 * a[i][1], sq(a[i][1]) + 2 * a[i][1] - sq(max(0ll, a[i][1] - a[i + 1][0] + 1)) + dp[i + 1][j - 1]);
            dp[i][j] = cur.query(a[i][0]) + sq(a[i][0]) - 2 * a[i][0] + 1;
        }
    }
    return dp[1][k];
}

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