oj.uz

답안 #201805

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
201805 2020-02-12T04:00:48 Z tri Bubble Sort 2 (JOI18_bubblesort2) C++14
0 / 100
 37 ms 3316 KB 
#include <bits/stdc++.h>
#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;

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;

#define pb push_back
#define f first
#define s second

// set
typedef tree<
        pi,
        null_type,
        less<pi>,
        rb_tree_tag,
        tree_order_statistics_node_update>
        ordered_set;

const int MAXN = 5e5 + 100;
const int INF = 1e9;

struct LSegTr {
    int tr[4 * MAXN], lz[4 * MAXN];

    void push(int i, int l, int r) {
        tr[i] += lz[i];
        if (l != r) {
            lz[i * 2] += lz[i];
            lz[i * 2 + 1] += lz[i];
        }
        lz[i] = 0;
    }

    int q(int i, int l, int r, int s, int e) {
        if (e < l || r < s) {
            return 0;
        }
        push(i, l, r);
        if (s <= l && r <= e) {
            return tr[i];
        }
        int mid = (l + r) / 2;
        return max(q(i * 2, l, mid, s, e), q(i * 2 + 1, mid + 1, r, s, e));
    }

    void u(int i, int l, int r, int s, int e, int d) {
        push(i, l, r); // pushed early to use in recalculation of parent
        if (e < l || r < s) {
            return;
        }
        if (s <= l && r <= e) {
            lz[i] += d;
            push(i, l, r);
            return;
        }
        int mid = (l + r) / 2;
        u(i * 2, l, mid, s, e, d);
        u(i * 2 + 1, mid + 1, r, s, e, d);
        tr[i] = max(tr[i * 2], tr[i * 2 + 1]);
    }

    void set(int i, int l, int r, int x, int v) {
        push(i, l, r);
        if (l == r) {
            tr[i] = v;
            return;
        }
        int mid = (l + r) / 2;
        if (x <= mid) {
            set(i * 2, l, mid, x, v);
        } else {
            set(i * 2 + 1, mid + 1, r, x, v);
        }
        tr[i] = max(tr[i * 2], tr[i * 2 + 1]);
    }

    void b(int i, int l, int r, vi &init) {
        lz[i] = 0;
        if (l == r) {
            tr[i] = init[l];
            return;
        }
        int mid = (l + r) / 2;
        b(i * 2, l, mid, init);
        b(i * 2 + 1, mid + 1, r, init);
    }

} lSegTr;


int N, Q;

ordered_set vals;

vi countScans(vi A, vi X, vi V) {
    vector<pi> allVal;

    vi a(A.begin(), A.end());
    N = a.size();
    Q = X.size();

    for (int i = 0; i < N; i++) {
        allVal.pb({a[i], i});
    }

    for (int i = 0; i < Q; i++) {
        allVal.pb({V[i], X[i]});
    }

    sort(allVal.begin(), allVal.end());
    allVal.erase(unique(allVal.begin(), allVal.end()), allVal.end());

    for (int i = 0; i < N; i++) {
        vals.insert({a[i], i});
    }

    int M = allVal.size();

    vi init(M);

    for (int cP = 0; cP < N; cP++) {
        pi cVal = {a[cP], cP};
        int gP = vals.order_of_key(cVal);

        int index = lower_bound(allVal.begin(), allVal.end(), cVal) - allVal.begin();
        init[index] = cP - gP;
    }

    lSegTr.b(1, 0, M - 1, init);

    vi ans(Q);
    for (int cQ = 0; cQ < Q; cQ++) {
        int cP = X[cQ];
        pi oV = {a[cP], cP};
        pi nV = {V[cQ], cP};

        a[cP] = V[cQ];
        vals.erase(oV);
        vals.insert(nV);

        int oIndex = upper_bound(allVal.begin(), allVal.end(), oV) - allVal.begin();
        int nIndex = upper_bound(allVal.begin(), allVal.end(), nV) - allVal.begin();

        if (oIndex < M) lSegTr.u(1, 0, M - 1, oIndex, M - 1, +1);

        if (nIndex < M) lSegTr.u(1, 0, M - 1, nIndex, M - 1, -1);

        lSegTr.set(1, 0, M - 1, oIndex, -INF);

        int gP = vals.order_of_key(nV);
        lSegTr.set(1, 0, M - 1, nIndex, cP - gP);

        ans[cQ] = lSegTr.q(1, 0, M - 1, 0, M - 1);
    }
    return ans;
}

Subtask #1
0 / 17.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 6 ms 376 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #2
0 / 21.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 6 ms 376 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #3
0 / 22.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 37 ms 3316 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #4
0 / 40.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 6 ms 376 KB Output isn't correct
2 Halted 0 ms 0 KB -