Submission #108393

#	Time	Username	Problem	Language	Result	Execution time	Memory
108393		Diuven	Split the sequence (APIO14_sequence)	C++14	100 / 100	996 ms	83452 KiB

sequence

#include <bits/stdc++.h>
using namespace std;

typedef long long lint;
typedef pair<int, int> pii;

const int INF = 2e9;
const int MOD = 1e9+7;
const int MAX = 1e5+10;
const lint LNF = 2e18;

int n, p;
lint S[MAX];
lint D[MAX], E[MAX];
int from[202][MAX];

struct frac{
	lint a, b; // a/b , b>0
	frac(lint x, lint y): a(x), b(y) {
		if(b==0){ cerr<<"ERROR!!\n", exit(42); }
		if(b<0) a*=-1, b*=-1;
	}
	frac(lint x): a(x), b(1) {}
	frac(): a(1), b(1) {}
	bool operator <= (const frac &f) {
		return a*f.b <= f.a*b;
	}
	bool operator <= (lint x) {
		return a <= x * b;
	}
	void show(){
		cout<<a<<"/"<<b<<" ~= "<<(double)a/b<<endl;
	}
};

struct line_t{
	lint a, b; int idx;
	lint eval(lint x){
		return a*x + b;
	}
	frac operator ^ (const line_t &l) const {
		return frac(l.b-b, a-l.a);
	}
	void show(){ cout<<a<<' '<<b<<endl; }
};
deque<line_t> Q;

/*
D[i][j]: 1~i까지 j조각 최대
D[i][j] = max(D[k][j-1] + (S[i]S[k]-S[k]^2)
		= max(S[k] * S[i] + (D[k][j-1]-S[k]^2))
*/

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

	cin>>n>>p; p++;
	for(int i=1, a; i<=n; i++) cin>>a, S[i]=S[i-1]+a;

	for(int i=1; i<=n; i++) D[i] =  -LNF;
	D[0] = 0;

	for(int j=1; j<=p; j++){
		Q.clear(); Q.push_back({0, D[0], 0});
		E[0] = -INF;
		for(int i=1; i<=n; i++){
			// cout<<"Solve: "<<i<<", "<<j<<endl;
			// for(line_t l:Q) l.show();
			// cout<<endl;

			while(Q.size() > 1U){
				// a,b가 S[i] 이하이면 pop_front
				line_t a = Q[0], b = Q[1];
				if((a^b) <= S[i]) Q.pop_front();
				else break;
			}

			E[i] = Q[0].eval(S[i]);
			from[j][i] = Q[0].idx;

			// cout<<"Midst\n";

			line_t now = {S[i], D[i] - S[i]*S[i], i};

			while(Q.size() > 1U){
				// a,b가 now,b가 이하이면 pop_back
				line_t a = Q.back(), b = Q[Q.size()-2U];
				if((now^b) <= (a^b)) Q.pop_back();
				else break;
			}
			// 만약 평행하다면, 절편이 더 큰걸로 교체
			if(!Q.empty() && Q.back().a == now.a){
				if(Q.back().b <= now.b)
					Q.back().b = now.b, Q.back().idx = now.idx;
			}
			else Q.push_back(now);
		}
		for(int i=0; i<=n; i++) D[i] = E[i];
	}

	int pos = -1; lint ans = -1;
	for(int i=1; i<=n; i++){
		if(ans<D[i]) pos = i, ans = D[i];
	}
	
	vector<int> track;
	for(int i=p; i>1; i--){
		pos = from[i][pos];
		track.push_back(pos);
	}

	cout<<ans<<'\n';
	reverse(track.begin(), track.end());
	for(int x:track) cout<<x<<' ';
	cout<<'\n';

	return 0;
}

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