Submission #155968

#	Time	Username	Problem	Language	Result	Execution time	Memory
155968		solarmagic	Split the sequence (APIO14_sequence)	C++17	0 / 100	2065 ms	83856 KiB

sequence

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define SZ(x) (int)(x).size()
#define ALL(x) (x).begin(),(x).end()
#define solarmagic ios_base::sync_with_stdio(0),cin.tie(0)

struct P { ll X, Y; int idx;};
struct CHT{
	vector<P> line;
	int now;

	CHT() { now = 0; }
	ll f(ll i, ld x) { return line[i].X*x + line[i].Y; }
	inline ld cross(int i, int j) { 
		return (ld)(line[i].Y - line[j].Y) / (ld)(line[j].X - line[i].X);
	}
	void insert(P a) {
		line.push_back(a);
		int i = SZ(line) - 1;

        // cht코드에 문제가 있다면 이 if문만 보면 될 듯(나머진 맞을거임)
        // (평행할 때 처리인데 이건 질문 게시판 나와 있는것처럼 평행하면 Y절편 이 높은걸 남김)
		if (i > 1) { 
			if (line[i].X == line[i-1].X) {
                if (line[i].Y > line[i-1].Y) line[i-1] = line[i];
				line.pop_back();
				return ;
			}
		}

		while (i > 1 && cross(i, i-1) < cross(i-1, i-2)) {
			line[i-1] = line.back();
			line.pop_back();
			i--;
		}
	}

	pair<ll,int> get(ll k) { 
		do {
			if (now == SZ(line)-1) return {f(now, k),line[now].idx};
			ll cr = cross(now, now+1);
			if (k < cr) return {f(now, k),line[now].idx};
			now++;
		} while (true);
	}

    void clear() {
        line.clear();
        now = 0;
    }
} cht;

ll s[100001];
ll d[100001][2];
vector<int> zeros;
int track[100001][201];
int main() {
    solarmagic;

    ll n, k; cin >> n >> k;
    for (int i = 1; i <= n; i++) {
    	cin >> s[i];
        if (s[i] == 0 && SZ(zeros) < k) zeros.push_back(i);
    	s[i] += s[i-1];
    }
    for (int j = 1; j <= k; j++) {
        cht.clear();
        cht.insert({0,0,-1});
    	for (int i = 1; i <= n; i++) {
    		auto p = cht.get(s[i]);
    		d[i][j%2] = p.first;
    		track[i][j] = p.second;
    		cht.insert({s[i],-s[i]*s[i]+d[i][1-j%2], i});
            //cout << track[i][j] << ' ';
    	}
        //cout << '\n';
    }	
    cout << d[n][k%2] << '\n'; //아마 여기까진 맞을듯


    // 트래킹할 때 0이 여러개 있으면 얘가 트래킹을 제대로 못해서 (ex: 6 4\n 2 1 0 0 0 0)
    // 일단 구한다음에 unique한것만 남기고
    // index가 zero인 얘들을 추가해주는 방식으로 함 (되나?)

    vector<int> ans;
    for (int i = k - 1; i >= 0; i--) {
    	n =  track[n][i+1];
        if (n == -1) break;
        ans.push_back(n);
    }
    sort(ALL(ans));
    ans.erase(unique(ALL(ans)), ans.end());
    for (int i = SZ(ans), j = 0; i < k; i++, j++) {
        ans.push_back(zeros[j]);
    }
    sort(ALL(ans));
    for (auto i : ans) cout << i << ' ';
    return 0;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 11
• Subtask 3: 0 / 11
• Subtask 4: 0 / 17
• Subtask 5: 0 / 21
• Subtask 6: 0 / 29