제출 #403961

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
403961		SavicS	선물상자 (IOI15_boxes)	C++14	50 / 100	54 ms	17444 KiB

boxes

#include "boxes.h"

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>

#define fi first
#define se second
#define pb push_back
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define ff(i,a,b) for(int i=a;i<=b;i++)
#define fb(i,b,a) for(int i=b;i>=a;i--)

using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 100005;
const ll inf = 1e18 + 5;

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

int n, k, l;
int niz[maxn];

ll dp1[maxn];
ll dp2[maxn];

int pola = 0;
ll dist1(int X, int Y){
	if(niz[Y] <= pola)return 2 * niz[Y];
	if(niz[X] <= pola)return l;
	else return 2 * niz[Y];
}

ll dist2(int X, int Y){
	if(niz[Y] >= pola)return 2 * (l - niz[Y]);
	if(niz[X] <= pola)return l;
	else return 2 * (l - niz[Y]);
}

ll delivery(int N, int K, int L, int positions[]){
	n = N, k = K, l = L;
	ff(i,1,n)niz[i] = positions[i - 1];
	
	ff(i,1,n)dp1[i] = dp2[i] = inf;
	pola = l / 2;

	dp1[0] = 0;
	ff(i,1,n){
		fb(j,i,max(1,i - k + 1)){
			dp1[i] = min(dp1[i], dp1[j - 1] + dist1(j, i));
		}
	}	
	
	dp2[n + 1] = 0;
	fb(i,n,1){
		ff(j,i,min(n, i + k - 1)){
			dp2[i] = min(dp2[i], dp2[j + 1] + dist2(j, i));
		}
	}
	
	ll rez = inf;
	ff(i,0,n)rez = min(rez, dp1[i] + dp2[i + 1]);
	
	return rez;
}

//int main()
//{
//    ios::sync_with_stdio(false);
//    cout.tie(nullptr);
//    cin.tie(nullptr);
//	cin >> n >> k >> l;
//	ff(i,1,n)cin >> niz[i];
//	
//	ff(i,1,n)dp1[i] = dp2[i] = inf;
//	pola = l / 2;
//
//	dp1[0] = 0;
//	ff(i,1,n){
//		fb(j,i,max(1,i - k + 1)){
//			dp1[i] = min(dp1[i], dp1[j - 1] + dist1(j, i));
//		}
//	}	
//	
//	dp2[n + 1] = 0;
//	fb(i,n,1){
//		ff(j,i,min(n, i + k - 1)){
//			dp2[i] = min(dp2[i], dp2[j + 1] + dist2(j, i));
//		}
//	}
//	
//	ll rez = inf;
//	ff(i,0,n)rez = min(rez, dp1[i] + dp2[i + 1]);
//
//	cout << rez << endl;
//		
//    return 0;
//}
/**

3 2 8
1 2 5

// probati bojenje sahovski ili slicno

**/

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