제출 #752922

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
752922		I_love_Hoang_Yen	서열 (APIO23_sequence)	C++17	7 / 100	308 ms	31832 KiB

sequence

#include "sequence.h"
#include <bits/stdc++.h>
#define SZ(s) ((int) ((s).size()))
using namespace std;

vector<int> medians(const vector<int> values) {
    int n = SZ(values);
    assert(n > 0);
    if (n % 2 == 0) {
        return {values[n/2 - 1], values[n / 2]};
    } else {
        return {values[n/2]};
    }
}

int sub1(const vector<int>& a) {
    int n = SZ(a);
    int res = 0;
    for (int l = 0; l < n; ++l) {
        for (int r = l; r < n; ++r) {
            vector<int> b(a.begin() + l, a.begin() + r + 1);

            unordered_map<int, int> cnt;
            for (int val : b) cnt[val] += 1;

            sort(b.begin(), b.end());
            auto meds = medians(b);
            for (int med : meds) {
                res = max(res, cnt[med]);
            }
        }
    }
    return res;
}

#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp>     // Including tree_order_statistics_node_update
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

using namespace __gnu_pbds;

int sub2(const vector<int>& a) {
    int n = SZ(a);
    int res = 0;
    for (int l = 0; l < n; ++l) {
        unordered_map<int, int> cnt;
        ordered_set values;
        for (int r = l; r < n; ++r) {
            cnt[a[r]]++;
            values.insert(a[r]);

            int k = SZ(values);
            if (k % 2 == 0) {
                res = max(res, cnt[*values.find_by_order(k / 2 - 1)]);
                res = max(res, cnt[*values.find_by_order(k / 2)]);
            } else {
                res = max(res, cnt[*values.find_by_order(k / 2)]);
            }
        }
    }
    return res;
}

bool is_sub_3(const std::vector<int> a) {
    auto mit = max_element(a.begin(), a.end());
    return std::is_sorted(a.begin(), mit)
        && std::is_sorted(mit, a.end(), std::greater<int>());
}

int len(const std::pair<int,int>& p) {
    return p.second - p.first + 1;
}
bool can_be_median(int cnt_less, int cnt_equal, int cnt_greater) {
    return cnt_equal + cnt_less >= cnt_greater;
}

int sub3(const vector<int>& a) {
    int n = SZ(a);
    unordered_map<int, vector<pair<int,int>>> pos;
    int l = 0;
    while (l < n) {
        int r = l;
        while (r < n && a[l] == a[r]) ++r;
        pos[a[l]].emplace_back(l, r-1);
        l = r;
    }
    int res = 0;
    for (const auto& [val, lrs] : pos) {
        // only 1 segment
        for (const auto& lr : lrs) res = max(res, len(lr));

        // 2 segments
        if (SZ(lrs) < 2) continue;
        assert(SZ(lrs) == 2);
        int cnt_equal = len(lrs[0]) + len(lrs[1]);
        int cnt_greater = len({lrs[0].second + 1, lrs[1].first - 1});
        int cnt_less = len({0, lrs[0].first - 1}) + len({lrs[1].second + 1, n-1});
        if (can_be_median(cnt_less, cnt_equal, cnt_greater)) {
            res = max(res, cnt_equal);
        }
    }
    return res;
}

// SegTree, copied from AtCoder library {{{
// AtCoder doc: https://atcoder.github.io/ac-library/master/document_en/segtree.html
//
// Notes:
// - Index of elements from 0 -> n-1
// - Range queries are [l, r-1]
//
// Tested:
// - (binary search) https://atcoder.jp/contests/practice2/tasks/practice2_j
// - https://oj.vnoi.info/problem/gss
// - https://oj.vnoi.info/problem/nklineup
// - (max_right & min_left for delete position queries) https://oj.vnoi.info/problem/segtree_itstr
// - https://judge.yosupo.jp/problem/point_add_range_sum
// - https://judge.yosupo.jp/problem/point_set_range_composite
int ceil_pow2(int n) {
    int x = 0;
    while ((1U << x) < (unsigned int)(n)) x++;
    return x;
}

template<
    class T,  // data type for nodes
    T (*op) (T, T),  // operator to combine 2 nodes
    T (*e)() // identity element
>
struct SegTree {
    SegTree() : SegTree(0) {}
    explicit SegTree(int n) : SegTree(vector<T> (n, e())) {}
    explicit SegTree(const vector<T>& v) : _n((int) v.size()) {
        log = ceil_pow2(_n);
        size = 1<<log;
        d = vector<T> (2*size, e());

        for (int i = 0; i < _n; i++) d[size+i] = v[i];
        for (int i = size - 1; i >= 1; i--) {
            update(i);
        }
    }

    // 0 <= p < n
    void set(int p, T x) {
        assert(0 <= p && p < _n);
        p += size;
        d[p] = x;
        for (int i = 1; i <= log; i++) update(p >> i);
    }

    // 0 <= p < n
    T get(int p) const {
        assert(0 <= p && p < _n);
        return d[p + size];
    }

    // Get product in range [l, r-1]
    // 0 <= l <= r <= n
    // For empty segment (l == r) -> return e()
    T prod(int l, int r) const {
        assert(0 <= l && l <= r && r <= _n);
        T sml = e(), smr = e();
        l += size;
        r += size;
        while (l < r) {
            if (l & 1) sml = op(sml, d[l++]);
            if (r & 1) smr = op(d[--r], smr);
            l >>= 1;
            r >>= 1;
        }
        return op(sml, smr);
    }

    T all_prod() const {
        return d[1];
    }

    // Binary search on SegTree to find largest r:
    //    f(op(a[l] .. a[r-1])) = true   (assuming empty array is always true)
    //    f(op(a[l] .. a[r])) = false    (assuming op(..., a[n]), which is out of bound, is always false)
    template <bool (*f)(T)> int max_right(int l) const {
        return max_right(l, [](T x) { return f(x); });
    }
    template <class F> int max_right(int l, F f) const {
        assert(0 <= l && l <= _n);
        assert(f(e()));
        if (l == _n) return _n;
        l += size;
        T sm = e();
        do {
            while (l % 2 == 0) l >>= 1;
            if (!f(op(sm, d[l]))) {
                while (l < size) {
                    l = (2 * l);
                    if (f(op(sm, d[l]))) {
                        sm = op(sm, d[l]);
                        l++;
                    }
                }
                return l - size;
            }
            sm = op(sm, d[l]);
            l++;
        } while ((l & -l) != l);
        return _n;
    }

    // Binary search on SegTree to find smallest l:
    //    f(op(a[l] .. a[r-1])) = true      (assuming empty array is always true)
    //    f(op(a[l-1] .. a[r-1])) = false   (assuming op(a[-1], ..), which is out of bound, is always false)
    template <bool (*f)(T)> int min_left(int r) const {
        return min_left(r, [](T x) { return f(x); });
    }
    template <class F> int min_left(int r, F f) const {
        assert(0 <= r && r <= _n);
        assert(f(e()));
        if (r == 0) return 0;
        r += size;
        T sm = e();
        do {
            r--;
            while (r > 1 && (r % 2)) r >>= 1;
            if (!f(op(d[r], sm))) {
                while (r < size) {
                    r = (2 * r + 1);
                    if (f(op(d[r], sm))) {
                        sm = op(d[r], sm);
                        r--;
                    }
                }
                return r + 1 - size;
            }
            sm = op(d[r], sm);
        } while ((r & -r) != r);
        return 0;
    }

private:
    int _n, size, log;
    vector<T> d;

    void update(int k) {
        d[k] = op(d[2*k], d[2*k+1]);
    }
};
// }}}
// SegTree examples {{{
// Examples: Commonly used SegTree ops: max / min / sum
struct MaxSegTreeOp {
    static int op(int x, int y) {
        return max(x, y);
    }
    static int e() {
        return INT_MIN;
    }
};

struct MinSegTreeOp {
    static int op(int x, int y) {
        return min(x, y);
    }
    static int e() {
        return INT_MAX;
    }
};

struct SumSegTreeOp {
    static long long op(long long x, long long y) {
        return x + y;
    }
    static long long e() {
        return 0;
    }
};

// using STMax = SegTree<int, MaxSegTreeOp::op, MaxSegTreeOp::e>;
// using STMin = SegTree<int, MinSegTreeOp::op, MinSegTreeOp::e>;
// using STSum = SegTree<int, SumSegTreeOp::op, SumSegTreeOp::e>;
// }}}

int sub4(const vector<int>& a) {
    int n = SZ(a);
    int res = 0;
    int ln = *max_element(a.begin(), a.end());
    for (int median = 0; median <= ln; ++median) {
        vector<int> cnt_less(n, 0);
        vector<int> cnt_equal(n, 0);
        vector<int> cnt_greater(n, 0);
        // [l, r] satisfies <=> cnt_less(l, r) + cnt_equal(l, r) - cnt_greater(l, r) >= 0
        // Fix l => Find max r:
        //     cnt_less(l, r) + cnt_equal(l, r) - cnt_greater(l, r) >= 0
        // <=> cnt_less[r] + cnt_equal[r] - cnt_greater[r]
        //       >= cnt_less[l-1] + cnt_equal[l-1] - cnt_greater[l-1]

        for (int i = 0; i < n; ++i) {
            cnt_less[i] = a[i] < median;
            cnt_equal[i] = a[i] == median;
            cnt_greater[i] = a[i] > median;
        }
        std::partial_sum(cnt_less.begin(), cnt_less.end(), cnt_less.begin());
        std::partial_sum(cnt_equal.begin(), cnt_equal.end(), cnt_equal.begin());
        std::partial_sum(cnt_greater.begin(), cnt_greater.end(), cnt_greater.begin());

        // st: stores max(r)
        SegTree<int, MaxSegTreeOp::op, MaxSegTreeOp::e> st(2 * n + 1);
        auto f = [&] (int i) {
            if (i < 0) return n;
            // +n to make sure result >= 0
            return cnt_less[i] + cnt_equal[i] - cnt_greater[i] + n;
        };
        for (int l = n-1; l >= 0; --l) {
            int r = st.prod(f(l-1), 2*n + 1);
            if (r >= l) {
                int equals = cnt_equal[r] - (l > 0 ? cnt_equal[l-1] : 0);
                res = max(res, equals);
            }
            st.set(f(l), max(st.get(f(l)), l));
        }
    }
    return res;
}

int sequence(int n, std::vector<int> a) {
    if (n <= 2000 || *max_element(a.begin(), a.end()) <= 3) return sub4(a);
    if (is_sub_3(a)) return sub3(a);
    return 0;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 17
• Subtask 3: 7 / 7
• Subtask 4: 0 / 12
• Subtask 5: 0 / 13
• Subtask 6: 0 / 22
• Subtask 7: 0 / 18