結果

問題 No.1247 ブロック登り
ユーザー 👑 hitonanodehitonanode
提出日時 2020-10-03 02:24:39
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 5,337 bytes
コンパイル時間 2,372 ms
コンパイル使用メモリ 211,824 KB
実行使用メモリ 4,508 KB
最終ジャッジ日時 2023-09-25 04:28:30
合計ジャッジ時間 24,397 ms
ジャッジサーバーID
（参考情報） 		judge12 / judge11
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
4,380 KB
testcase_01 AC 2 ms
4,380 KB
testcase_02 AC 2 ms
4,380 KB
testcase_03 AC 2 ms
4,380 KB
testcase_04 WA -
testcase_05 WA -
testcase_06 AC 2 ms
4,376 KB
testcase_07 WA -
testcase_08 WA -
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 WA -
testcase_20 WA -
testcase_21 AC 3 ms
4,376 KB
testcase_22 WA -
testcase_23 AC 4 ms
4,376 KB
testcase_24 AC 2 ms
4,380 KB
testcase_25 AC 1,506 ms
4,380 KB
testcase_26 WA -
testcase_27 WA -
testcase_28 WA -
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
using namespace std;
using lint = long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios {
    fast_ios() { cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(20); };
} fast_ios_;
#define FOR(i, begin, end) for (int i = (begin), i##_end_ = (end); i < i##_end_; i++)
#define IFOR(i, begin, end) for (int i = (end)-1, i##_begin_ = (begin); i >= i##_begin_; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define ALL(x) (x).begin(), (x).end()
//
template <typename T, typename V>
void ndarray(vector<T>& vec, const V& val, int len) { vec.assign(len, val); }
template <typename T, typename V, typename... Args>
void ndarray(vector<T>& vec, const V& val, int len, Args... args)
{
    vec.resize(len), for_each(begin(vec), end(vec), [&](T& v) { ndarray(v, val, args...); });
}
template <typename T>
bool chmax(T& m, const T q) { return m < q ? (m = q, true) : false; }
template <typename T>
bool chmin(T& m, const T q) { return m > q ? (m = q, true) : false; }
template <typename T1, typename T2>
pair<T1, T2> operator+(const pair<T1, T2>& l, const pair<T1, T2>& r) { return make_pair(l.first + r.first, l.second + r.second); }
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2>& l, const pair<T1, T2>& r) { return make_pair(l.first - r.first, l.second - r.second); }
template <typename T>
vector<T> srtunq(vector<T> vec) { return sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()), vec; }
template <typename T>
istream& operator>>(istream& is, vector<T>& vec)
{
    return for_each(begin(vec), end(vec), [&](T& v) { is >> v; }), is;
}

// output
template <typename T, typename V>
ostream& dmpseq(ostream&, const T&, const string&, const string&, const string&);
#if __cplusplus >= 201703L
template <typename... T>
ostream& operator<<(ostream& os, const tuple<T...>& tpl)
{
    return apply([&os](auto&&... args) { ((os << args << ','), ...); }, tpl), os;
}
#endif
//
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) { return os << '(' << p.first << ',' << p.second << ')'; }
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& x) { return dmpseq<vector<T>, T>(os, x, "[", ",", "]"); }
template <typename T>
ostream& operator<<(ostream& os, const deque<T>& x) { return dmpseq<deque<T>, T>(os, x, "deq[", ",", "]"); }
template <typename T>
ostream& operator<<(ostream& os, const set<T>& x) { return dmpseq<set<T>, T>(os, x, "{", ",", "}"); }
template <typename T, typename TH>
ostream& operator<<(ostream& os, const unordered_set<T, TH>& x) { return dmpseq<unordered_set<T, TH>, T>(os, x, "{", ",", "}"); }
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& x) { return dmpseq<multiset<T>, T>(os, x, "{", ",", "}"); }
template <typename TK, typename T>
ostream& operator<<(ostream& os, const map<TK, T>& x) { return dmpseq<map<TK, T>, pair<TK, T>>(os, x, "{", ",", "}"); }
template <typename TK, typename T, typename TH>
ostream& operator<<(ostream& os, const unordered_map<TK, T, TH>& x) { return dmpseq<unordered_map<TK, T, TH>, pair<TK, T>>(os, x, "{", ",", "}"); }
template <typename T, typename V>
ostream& dmpseq(ostream& os, const T& seq, const string& pre, const string& sp, const string& suf)
{
    return os << pre, for_each(begin(seq), end(seq), [&](V x) { os << x << sp; }), os << suf;
}
template <typename T>
void print(const vector<T>& x) { dmpseq<vector<T>, T>(cout, x, "", " ", "\n"); }
#ifdef HITONANODE_LOCAL
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl
#else
#define dbg(x) {}
#endif

vector<int> solve(int N, int K, const vector<int> A)
{
    vector<int> ret(N, -2e9);
    REP(i, N)
    {
        int cost = K * A[i];
        FOR(j, i + 1, N)
        {
            int h = K - (j - i);
            if (h < 0) continue;

            cost += h * A[j];
            int tmp = cost;
            vector<vector<int>> Begin(N), End(N);
            int l = max(i, j - h), r = j;
            if (abs(i - l) % 2 != K % 2) l++;
            if (abs(i - r) % 2 != K % 2) r--;
            Begin[l].emplace_back(tmp), End[r].emplace_back(tmp);

            IREP(k, i)
            {
                int h = K - (j - i) - (j - k);
                if (h < 0) break;
                tmp += h * A[k];
                int l = k, r = min(k + h, j);
                if (abs(i - l) % 2 != K % 2) l++;
                if (abs(i - r) % 2 != K % 2) r--;
                Begin[l].emplace_back(tmp), End[r].emplace_back(tmp);
            }

            vector<multiset<int>> ms(2);
            REP(i, N)
            {
                ms[i % 2].insert(Begin[i].begin(), Begin[i].end());
                if (ms[i % 2].size()) chmax(ret[i], *prev(ms[i % 2].end()));
                for (auto x : End[i])
                {
                    ms[i % 2].erase(ms[i % 2].lower_bound(x));
                }
            }
        }
    }
    return ret;
}

int main()
{
    int N, K;
    cin >> N >> K;
    vector<int> A(N);
    cin >> A;
    vector<int> ret1 = solve(N, K, A);
    reverse(A.begin(), A.end());
    vector<int> ret2 = solve(N, K, A);
    reverse(ret2.begin(), ret2.end());

    REP(i, N)
    {
        cout << max(ret1[i], ret2[i]) << '\n';
    }
}
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