Submission #222160

#	Time	Username	Problem	Language	Result	Execution time	Memory
222160		VEGAnn	새로운 문제 (COCI19_akvizna)	C++14	65 / 130	1099 ms	262144 KiB

Main

#include <bits/stdc++.h>
#define sz(x) ((int)x.size())
#define pii pair<int,int>
#define ft first
#define sd second
#define MP make_pair
#define all(x) x.begin(),x.end()
#define PB push_back
using namespace std;
typedef long long ll;
typedef long double ld;
const int N = 3010;
const ll OO = 1e18;
const ld E = 1e-9;
vector<int> vc;
ld f[N][N];
int n, k;

struct line{
    ld k, b;

    line(): k(0.0), b(0.0) {}
    line(ld _k, ld _b): k(_k), b(_b) {}
};

deque<line> hull;

ld get_cross_point(line fi, line se){
    return (fi.b - se.b) / (se.k - fi.k);
}

void insert(line nw){
    if (sz(hull) == 0){
        hull.PB(nw);
        return;
    }

    while (sz(hull) > 1){
        ld pt1 = get_cross_point(nw, hull[1]);
        ld pt2 = get_cross_point(hull.back(), hull[0]);

        if (pt2 > pt1) break;

        hull.pop_front();
    }

    hull.push_front(nw);
}

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);

//    freopen("in.txt","r",stdin);

    cin >> n >> k;

    for (int i = 0; i <= k; i++)
        for (int j = 0; j <= n; j++)
            f[i][j] = -1.0;

    f[0][n] = 0.0;

    for (int i = 1; i <= k; i++) {

        hull.clear();

        if (f[i - 1][n] >= -E)
            insert({-1.0 / ld(n), f[i - 1][n] + 1.0});

        for (int j = n - 1; j >= 0; j--){

            if (sz(hull)) {

                int l = 0, r = sz(hull) - 1;

                while (l < r) {
                    int md = (l + r + 1) >> 1;

                    assert(md > 0);

                    if (get_cross_point(hull[md], hull[md - 1]) + E <= j)
                        l = md;
                    else r = md - 1;
                }

                f[i][j] = hull[l].b + hull[l].k * ld(j);
            }

            // may be this should be after update f[i][j]
            if (f[i - 1][j] >= -E)
                insert({-1.0 / ld(j), f[i - 1][j] + 1.0});
        }
    }

    cout << fixed << setprecision(10) << f[k][0];

    return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 5 / 5
• Subtask 3: 5 / 5
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