제출 #127993

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
127993		MoNsTeR_CuBe	선물상자 (IOI15_boxes)	C++17	10 / 100	170 ms	828 KiB

boxes

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

long long delivery(int N, int K, int L, int p[]) {
	#define int long long 
	
	if(N <= 10){
		
		vector< int > v(N);
		for(int i = 0; i < N; i++){
			v[i] = p[i];
		}
		
		int ans = numeric_limits<int>::max();
		
		do{
			
			for(int a : v) cout << a << ' ';
			cout << endl;
			
			int last = 0;
			int curr = K;
			
			int cost = 0;
			
			for(int i = 0; i < (int) v.size(); i++){
				if(abs(v[i]-last) + (curr ? 0 : min(last, abs(L-last))) < min(abs(L-v[i]), v[i]) + min(abs(L-last), last)){
					if(!curr){
						curr = K;
						cost += min(last, abs(L-last));
					}
					curr--;
					cost += abs(v[i]-last);
				}else{
					curr = K-1;
					cost += min(abs(L-v[i]), v[i]) + min(abs(L-last), last);
				}
				
				cout << "STEP " << cost << endl;
				
				last = v[i];
			}
			
			ans = min(ans, cost + min(last, L-last));
			
			cout << "COST " << cost + min(last, L-last) << endl;
			
		}while(next_permutation(v.begin(), v.end()));
		
		return ans;
	}
	
	vector< int > left;
	
	int curr = K;
	
	int last = 0;
	
	int currCost = 0;
	
	for(int i = 0; i < N; i++){
		left.push_back(p[i]-last + currCost);
		curr--;
		if(curr == 0){
			curr = K;
			currCost += 2*p[i];
		}
		currCost += p[i]-last;
		last = p[i];
	}
	//left.push_back(L-last);
	/*for(int a : left) cout << a << ' ';
	cout << endl;*/
	
	vector< int > right;
	
	curr = K;
	last = L;
	currCost = 0;
	
	for(int i = N-1; i >= 0; i--){
		right.push_back(last - p[i] + currCost);
		curr--;
		if(curr == 0){
			curr = K;
			currCost += 2*(L-p[i]);
		}
		currCost += last - p[i];
		last = p[i];
	}
	//right.push_back(last + currCost);
	/*for(int a : right) cout << a << ' ';
	cout << endl;*/
	
	//reverse(right.begin(), right.end());
	
	int ans = min(right[N - 1] + min(p[0], L-p[0]), left[N-1] + min(L-p[N-1], p[N-1]));
	
	for(int i = 0; i < N; i++){
		if(N - 2 - i < 0) continue;
		//cout << "INDEX " << i << ' ' << left[i] << ' ' << right[N-2-i] << ' ' << min(p[i], L-p[i]) << ' ' << min(p[i+1], L-p[i+1]) << endl;
		//cout << p[N-2-i] << ' ' << L-p[N-2-i] << endl;
		ans = min(ans, left[i] + right[N - 2 - i] + min(p[i],L-p[i]) + min(p[i+1],L-p[i+1]));
	}
	
	return ans;
	#undef int
}

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