제출 #711340

#제출 시각아이디문제언어결과실행 시간메모리
711340Nhoksocqt1선물상자 (IOI15_boxes)C++17
100 / 100
764 ms336016 KiB

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

#define inf 0x3f3f3f3f
#define sz(x) int((x).size())
#define fi first
#define se second
typedef long long ll;
typedef pair<int, int> ii;

template<class X, class Y>
	inline bool maximize(X &x, const Y &y) {return (x < y ? x = y, 1 : 0);}
template<class X, class Y>
	inline bool minimize(X &x, const Y &y) {return (x > y ? x = y, 1 : 0);}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int Random(int l, int r) {
    return uniform_int_distribution<int>(l, r)(rng);
}

const int MAXN = 1e7+7;

ll dpL[MAXN], dpR[MAXN];
int a[MAXN], nArr, k, L;

#ifdef Nhoksocqt1

ll delivery(int _N, int _K, int _L, int _pos[]);

int _P[MAXN];

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    freopen("BOXES.inp", "r", stdin);
    freopen("BOXES.out", "w", stdout);

    int N, K, L;
    cin >> N >> K >> L;
    //N = Random(1, 10), K = Random(1, N), L = Random(1, 1e9); cout << N << ' ' << K << ' ' << L << '\n';
    for (int i = 0; i < N; ++i)
        cin >> _P[i];

    //for (int i = 0; i < N; ++i) _P[i] = Random(0, L - 1); sort(_P, _P + N); for (int i = 0; i < N; ++i) cout << _P[i] << " \n"[i + 1 == N];

    cout << delivery(N, K, L, _P) << '\n';
    return 0;
}

#endif // Nhoksocqt1

ll delivery(int _N, int _K, int _L, int _pos[]) {
    nArr = _N, k = _K, L = _L;

    int cnt0(0);
    while(cnt0 < nArr && _pos[cnt0] == 0)
        ++cnt0;

    for (int i = cnt0 + 1; i <= nArr; ++i)
        a[i - cnt0] = _pos[i - 1];

    nArr = _N - cnt0;

    for (int i = 0; i <= nArr + 1; ++i)
        dpL[i] = dpR[i] = 1e18;

    dpL[0] = dpR[nArr + 1] = 0;

    deque<int> dq;
    dq.push_back(0);

    for (int i = 1; i <= nArr; ++i) {
        dpL[i] = dpL[dq.front()] + min(L, 2 * a[i]);
        while(dq.size() && dpL[dq.back()] >= dpL[i])
            dq.pop_back();

        dq.push_back(i);
        if(dq.front() + k <= i)
            dq.pop_front();
    }

    while(dq.size())
        dq.pop_back();

    dq.push_back(nArr + 1);
    for (int i = nArr; i > 0; --i) {
        dpR[i] = dpR[dq.front()] + min(L, 2 * (L - a[i]));
        while(dq.size() && dpR[dq.back()] >= dpR[i])
            dq.pop_back();

        dq.push_back(i);
        if(dq.front() - k >= i)
            dq.pop_front();
    }

    ll result(1e18);
    for (int i = 0; i <= nArr; ++i)
        result = min(result, dpL[i] + dpR[i + 1]);

    return result;
}
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