結果

問題 No.2665 Minimize Inversions of Deque
ユーザー k1suxuk1suxu
提出日時 2024-03-08 22:07:23
言語 C++23
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 7,638 bytes
コンパイル時間 3,465 ms
コンパイル使用メモリ 264,072 KB
実行使用メモリ 17,216 KB
最終ジャッジ日時 2024-09-29 19:42:00
合計ジャッジ時間 11,099 ms
ジャッジサーバーID
(参考情報)
judge5 / judge2
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
6,816 KB
testcase_01 WA -
testcase_02 WA -
testcase_03 AC 135 ms
6,816 KB
testcase_04 AC 136 ms
6,816 KB
testcase_05 AC 138 ms
6,816 KB
testcase_06 WA -
testcase_07 AC 138 ms
6,816 KB
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 WA -
testcase_22 WA -
testcase_23 WA -
testcase_24 WA -
testcase_25 WA -
testcase_26 WA -
testcase_27 WA -
testcase_28 WA -
testcase_29 WA -
testcase_30 WA -
testcase_31 WA -
testcase_32 WA -
testcase_33 WA -
testcase_34 WA -
testcase_35 WA -
testcase_36 AC 260 ms
16,172 KB
testcase_37 WA -
testcase_38 WA -
testcase_39 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 #

// #pragma GCC target("avx")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")

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

#define rep(i,n) for(int i = 0; i < (int)n; i++)
#define FOR(n) for(int i = 0; i < (int)n; i++)
#define repi(i,a,b) for(int i = (int)a; i < (int)b; i++)
#define all(x) x.begin(),x.end()
//#define mp make_pair
#define vi vector<int>
#define vvi vector<vi>
#define vvvi vector<vvi>
#define vvvvi vector<vvvi>
#define pii pair<int,int>
#define vpii vector<pair<int,int>>

template<typename T>
bool chmax(T &a, const T b) {if(a<b) {a=b; return true;} else {return false;}}
template<typename T>
bool chmin(T &a, const T b) {if(a>b) {a=b; return true;} else {return false;}}

using ll = long long;
using ld = long double;
using ull = unsigned long long;

const ll INF = numeric_limits<long long>::max() / 2;
const ld pi = 3.1415926535897932384626433832795028;
const ll mod = 998244353;
int dx[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};

#define int long long

struct fenwick_tree {
    int n, sz;
    vector<int> dat;
    function<int(int, int)> f = [](int x, int y) {return x + y;};

    fenwick_tree(int n_) : dat(4*n_, 0), sz(n_) {
        int x = 1;
        while(x < n_) x *= 2;
        n = x;
    }

    void add(int i, int x) {
        i += n - 1;
        dat[i] += x;
        while(i > 0) {
            i = (i - 1) / 2;
            dat[i] = f(dat[i * 2 + 1], dat[i * 2 + 2]);
        }
    }

    void insert(int i) {
        return add(i, 1);
    }

    void erase(int i) {
        return add(i, -1);
    }

    void update(int i, int x) {
        i += n - 1;
        dat[i] = x;
        while(i > 0) {
            i = (i - 1) / 2;
            dat[i] = f(dat[i * 2 + 1], dat[i * 2 + 2]);
        }
    }

    int get_sum(int a, int b) {
        return get_sum_sub(a, b, 0, 0, n);
    }
    int get_sum_sub(int a, int b, int k, int l, int r) {
        if(r <= a || b <= l) {
            return 0;
        }else if(a <= l && r <= b) {
            return dat[k];
        }else {
            int vl = get_sum_sub(a, b, k * 2 + 1, l, (l + r) / 2);
            int vr = get_sum_sub(a, b, k * 2 + 2, (l + r) / 2, r);
            return f(vl, vr);
        }
    }

    int get(int i) {
        return dat[i + n - 1];
    }

    inline void print() {
        cout << "{ ";
        for(int i = 0; i < sz; i++) {
            cout << dat[i + n - 1] << " ";
        }
        cout << "}\n";
    }
};

namespace internal {
    // @param n `0 <= n`
    // @return minimum non-negative `x` s.t. `n <= 2**x`
    int ceil_pow2(int n) {
        int x = 0;
        while ((1U << x) < (unsigned int)(n)) x++;
        return x;
    }

    // @param n `1 <= n`
    // @return minimum non-negative `x` s.t. `(n & (1 << x)) != 0`
    constexpr int bsf_constexpr(unsigned int n) {
        int x = 0;
        while (!(n & (1 << x))) x++;
        return x;
    }

    // @param n `1 <= n`
    // @return minimum non-negative `x` s.t. `(n & (1 << x)) != 0`
    int bsf(unsigned int n) {
    #ifdef _MSC_VER
        unsigned long index;
        _BitScanForward(&index, n);
        return index;
    #else
        return __builtin_ctz(n);
    #endif
    }
}  // namespace internal

template <class S, S (*op)(S, S), S (*e)()> struct segtree {
  public:
    segtree() : segtree(0) {}
    explicit segtree(int n) : segtree(std::vector<S>(n, e())) {}
    explicit segtree(const std::vector<S>& v) : _n((int)v.size()) {
        log = internal::ceil_pow2(_n);
        size = 1 << log;
        d = std::vector<S>(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);
        }
    }

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

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

    S prod(int l, int r) const {
        assert(0 <= l && l <= r && r <= _n);
        S 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);
    }

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

    template <bool (*f)(S)> int max_right(int l) const {
        return max_right(l, [](S 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;
        S 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;
    }

    template <bool (*f)(S)> int min_left(int r) const {
        return min_left(r, [](S 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;
        S 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;
    std::vector<S> d;

    void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }
};

using S = int;
S op(S x, S y) {
    return x + y;
}
S e() {
    return 0;
}

template <typename T>
vector<T> compress(vector<T> &X) {
    vector<T> vals = X;
    sort(vals.begin(), vals.end());
    vals.erase(unique(vals.begin(), vals.end()), vals.end());
    for (int i = 0; i < (int)X.size(); i++) {
        X[i] = lower_bound(vals.begin(), vals.end(), X[i]) - vals.begin();
    }
    return vals;
}

int the_number_of_inversions(vector<long long> a) {
    int n = (int)a.size();
    compress<long long>(a);

    segtree<S, op, e> seg(n);
    int inversion = 0;

    FOR(n) {
        inversion += seg.prod(a[i]+1, n);
        seg.set(a[i], seg.get(a[i]) + 1);
    }

    return inversion;
}

void solve() {
    int n;
    cin >> n;
    vi p(n);
    FOR(n) {
        cin >> p[i];
        --p[i];
    }
    int inv = the_number_of_inversions(p);
    fenwick_tree fw(n);
    FOR(n) fw.insert(p[i]);
    vi pref, suff;
    for(int i = n-1; i >= 0; i--) {
        fw.erase(p[i]);
        if(fw.get_sum(0, p[i])-fw.get_sum(p[i],  n) < 0 || (fw.get_sum(0, p[i])-fw.get_sum(p[i],  n) == 0 && p[i] < p[0])) {
            pref.push_back(p[i]);
            inv += fw.get_sum(0, p[i])-fw.get_sum(p[i],  n);
        }else {
            suff.push_back(p[i]);
        }
    }
    reverse(all(suff));
    vi ans;
    for(auto e : pref) ans.push_back(e);
    for(auto e : suff) ans.push_back(e);

    cout << inv << endl;
    for(auto e : ans) cout << e+1 << " "; cout << endl;
}

signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}
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