제출 #296896

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
296896		Hemimor	선물상자 (IOI15_boxes)	C++14	70 / 100	2036 ms	282712 KiB

boxes

#include "boxes.h"
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <vector>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<P,int> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

template <class T>
class RMQ{
	public:
	int n;
	vector<T> rmq;
	T dfs(int a,int b,int k,int l,int r){
		if(r<=a||b<=l) return INF;
		if(a<=l&&r<=b) return rmq[k];
		int m=(l+r)/2;
		return min(dfs(a,b,k*2+1,l,m),dfs(a,b,k*2+2,m,r));
	}
	RMQ(int n_){
		n=1;
		while(n<n_) n*=2;
		rmq=vector<T>(2*n-1,INF);
	}
	void Update(int k,T x){
		k+=n-1;
		rmq[k]=x;
		while(k>0){
			k=(k-1)/2;
			rmq[k]=min(rmq[k*2+1],rmq[k*2+2]);
		}
	}
	T Query(int a,int b){
		return dfs(a,b,0,0,n);
	}
};

ll delivery(int N, int K, int L, int p[]) {
	vl c;
	sort(p,p+N);
	for(int i=0;i<N;i++) if(p[i]) c.push_back(p[i]);
	N=(int)c.size();
	vl a(N+1),b(N+1); 
	for(int i=0;i<N;i++) a[i+1]=(i+1-K>=0?a[i+1-K]:0)+c[i];
	for(int i=N-1;i>=0;i--) b[N-i]=(N-i-K>=0?b[N-i-K]:0)+L-c[i];
	ll res=INF;
	if(K<=4000){
	RMQ<ll> rmq(K);
	vl d(K,INF);
	for(int i=0;i<=N;i++){
		d[(i+K-1)%K]+=L;
		rmq.Update((i+K-1)%K,d[(i+K-1)%K]);
		if(a[i]*2<d[i%K]){
			d[i%K]=a[i]*2;
			rmq.Update(i%K,d[i%K]);
		}
		res=min(res,b[N-i]*2+rmq.rmq[0]);
	}
	}
	else{
		auto f=[&](ll i,ll j){
			return a[i]*2+b[j]*2+(ll)(N-i-j+K-1)/K*L;
		};
		for(int i=1;i<=N;i++) res=min({res,f(i,0),f(0,i),f(i,N-i)});
		for(int i=1;i<=N;i++) for(int j=i+K;j<=N;j+=K) res=min(res,f(i,N-j));
	}
	return res;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 15 / 15
• Subtask 4: 15 / 15
• Subtask 5: 20 / 20
• Subtask 6: 0 / 30