Submission #18459

#	Time	Username	Problem	Language	Result	Execution time	Memory
18459		tlwpdus	Split the sequence (APIO14_sequence)	C++98	100 / 100	835 ms	88200 KiB

sequence

/*
O(NK) ....?
*/

#include<stdio.h>
#include<algorithm>

using namespace std;

typedef long long ll;

struct fr { // p/q -> 진짜로 만들어야만 했을까
	ll p, q;
	fr(ll p = 0, ll q = 1):p(p),q(q){}
	inline bool operator < (const fr &A) const {return (q*A.q>0)?(p*A.q<q*A.p):(p*A.q>q*A.p);}
	inline bool operator == (const fr &A) const {return p*A.q==q*A.p;}
	inline bool operator <= (const fr &A) const {return (*this)<A||(*this)==A;}
	inline bool operator > (const fr &A) const {return !((*this)<=A);}
	inline bool operator < (const ll &A) const {return (*this)<fr(A,1);}
	inline bool operator == (const ll &A) const {return (*this)==fr(A,1);}
	inline bool operator > (const ll &A) const {return (*this)>fr(A,1);}
};

struct line {
	ll a, b;
	int idx;
	line(ll a=0, ll b=0, int idx=0):a(a),b(b),idx(idx){}
	fr operator * (const line &A) const {return fr(b-A.b,A.a-a);}
};

int n, m;
int arr[100010];
ll psum[100010];
ll dyn[2][100010];
int via[210][100010];

line st[100010];
int key = 0;

void pop(){key--;}
void push(line a) {
	while(key>1) {
		if (st[key-1].a==a.a) {
			if (st[key-1].b<a.b) pop();
			else return;
			continue;
		}
		if (st[key-1]*st[key-2]<=st[key-2]*a) pop();
		else break;
	}
	st[key++] = a;
}
int s;
pair<ll,int> getval(ll a) {
	int e = key-2, m;
	s = max(min(s,key-2),0);
	while(s<=e) {
		if (st[s]*st[s+1]<=a) return pair<ll,int>((a*st[s].a+st[s].b),st[s].idx);
		s++;
	}
	return pair<ll,int>(a*st[s].a+st[s].b,st[s].idx);
}
void clear() {key=0;}

void process() {
	int i, j, k;
	for (i=1;i<=n;i++) dyn[0][i] = psum[i]*(psum[n]-psum[i]);
	for (i=2;i<=m;i++) {
		clear();
		for (j=1;j<i;j++) {push(line(-psum[j],dyn[0][j],j));dyn[1][j]=0;}
		for (j=i;j<n;j++) {
			pair<ll,int> pli = getval(psum[n]-psum[j]);
			dyn[1][j] = pli.first+psum[j]*(psum[n]-psum[j]);
			via[i][j] = pli.second;
			push(line(-psum[j],dyn[0][j],j));
		}
		for (j=1;j<=n;j++) dyn[0][j] = dyn[1][j];
	}
	ll maxi = 0;
	int idx = -1;
	for (i=1;i<n;i++) {
		if (maxi<=dyn[0][i]) {
			maxi = dyn[0][i];
			idx = i;
		}
	}
	printf("%lld\n",maxi);
	for (i=m;i>0;idx=via[i--][idx]) printf("%d ",idx);
	printf("\n");
}

void input() {
	int i;
	scanf("%d %d",&n,&m);
	for (i=1;i<=n;i++) {
		scanf("%d",&arr[i]);
		psum[i] = psum[i-1]+arr[i];
	}
}

int main() {
	input();
	process();
	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