제출 #550447

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
550447		elazarkoren	선물상자 (IOI15_boxes)	C++17	50 / 100	2076 ms	12928 KiB

boxes

#include <bits/stdc++.h>
#include "boxes.h"
#define x first
#define y second
#define all(v) v.begin(), v.end()
#define chkmin(a, b) a = min(a, b)
#define chkmax(a, b) a = max(a, b)
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pii;
typedef vector<pii> vii;

const int MAX_N = 1e6 + 5;
const ll infinity = 1e18;

//ll dp[MAX_N];

//ll Solve(vi &p, int l) {
//    int n = p.size();
//    ll ans = l;
//    sort(all(p));
//    for (int i = 0; i < n - 1; i++) {
//        chkmin(ans, ll(2) * (p[i] + (l - p[i + 1])));
//    }
//    chkmin(ans, ll(2) * p[n - 1]);
//    chkmin(ans, ll(2) * (l - p[0]));
//    return ans;
//}

int n, k, l;

ll dpl[MAX_N], dpr[MAX_N];

void CalcL(vi &p) {
    for (int i = 1; i <= n; i++) {
        dpl[i] = dpl[i - 1];
        int m = max(0, i - k);
        for (int j = i - 1; j >= m; j--) {
            chkmin(dpl[i], dpl[j]);
        }
        dpl[i] += min<ll>(l, 2 * p[i]);
    }
}

void CalcR(vi &p) {
    for (int i = n; i >= 0; i--) {
        dpr[i] = dpr[i + 1];
        int m = min(n + 1, i + k);
        for (int j = i + 1; j <= m; j++) {
            chkmin(dpr[i], dpr[j]);
        }
        dpr[i] += min<ll>(l, 2 * (l - p[i]));
    }
}

inline ll Solve(int ind) {
    return dpl[ind - 1] + dpr[ind];
}

long long delivery(int N, int K, int L, int P[]) {
    n = N, k = K, l = L;
    vi p(n + 1);
    ll ans = infinity;
    for (int i = 1; i <= n; i++) p[i] = P[i - 1];
    CalcL(p), CalcR(p);
    for (int ind = 1; ind <= n + 1; ind++) {
        chkmin(ans, Solve(ind));
    }
//    sort(p, p + n);
//    for (int i = 0; i < n; i++) {
//        vi curr;
//        dp[i] = infinity;
//        for (int j = i; j >= 0; j--) {
//            curr.push_back(p[j]);
//            if (curr.size() > k) break;
//            ll x = (j > 0 ? dp[j - 1] : 0);
//            chkmin(dp[i], x + Solve(curr, l));
//        }
//    }
//    chkmin(ans, dp[n - 1]);
    return ans;
}
//3 2 8
//1 2 5

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