Submission #835448

#	Time	Username	Problem	Language	Result	Execution time	Memory
835448		SamAnd	Radio Towers (IOI22_towers)	C++17	60 / 100	1126 ms	74972 KiB

towers

#include "towers.h"
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define m_p make_pair
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
typedef long long ll;
const int N = 100005;

int n, d;
int a[N];

int ul[N], ur[N];
vector<pair<pair<int, int>, pair<int, int> > > b;

int z;
int s[N * 18 * 18];
int sl[N * 18 * 18], sr[N * 18 * 18];
void ubdy(int tl, int tr, int l, int r, int posy)
{
    if (l > r)
        return;
    if (tl == l && tr == r)
    {
        s[posy]++;
        return;
    }
    int m = (tl + tr) / 2;
    if (sl[posy] == 0)
        sl[posy] = ++z;
    if (sr[posy] == 0)
        sr[posy] = ++z;
    ubdy(tl, m, l, min(m, r), sl[posy]);
    ubdy(m + 1, tr, max(m + 1, l), r, sr[posy]);
}

int r[N * 4];
void ubdx(int tl, int tr, int lx, int rx, int ly, int ry, int posx)
{
    if (lx > rx)
        return;
    if (tl == lx && tr == rx)
    {
        if (r[posx] == 0)
            r[posx] = ++z;
        ubdy(1, n, ly, ry, r[posx]);
        return;
    }
    int m = (tl + tr) / 2;
    ubdx(tl, m, lx, min(m, rx), ly, ry, posx * 2);
    ubdx(m + 1, tr, max(m + 1, lx), rx, ly, ry, posx * 2 + 1);
}

int qryy(int tl, int tr, int y, int posy)
{
    if (posy == 0)
        return 0;
    if (tl == tr)
        return s[posy];
    int m = (tl + tr) / 2;
    if (y <= m)
        return qryy(tl, m, y, sl[posy]) + s[posy];
    return qryy(m + 1, tr, y, sr[posy]) + s[posy];
}

int qryx(int tl, int tr, int x, int y, int posx)
{
    if (tl == tr)
        return qryy(1, n, y, r[posx]);
    int m = (tl + tr) / 2;
    if (x <= m)
        return qryx(tl, m, x, y, posx * 2) + qryy(1, n, y, r[posx]);
    return qryx(m + 1, tr, x, y, posx * 2 + 1) + qryy(1, n, y, r[posx]);
}

void pre()
{
    b.clear();

    vector<int> s;
    for (int i = 1; i <= n; ++i)
    {
        while (!s.empty() && a[s.back()] <= a[i])
            s.pop_back();
        s.push_back(i);

        int l = 0, r = sz(s) - 1;
        ul[i] = 0;
        while (l <= r)
        {
            int m = (l + r) / 2;
            if (a[s[m]] - d >= a[i])
            {
                ul[i] = s[m];
                l = m + 1;
            }
            else
            {
                r = m - 1;
            }
        }
    }
    s.clear();
    for (int i = n; i >= 1; --i)
    {
        while (!s.empty() && a[s.back()] <= a[i])
            s.pop_back();
        s.push_back(i);

        int l = 0, r = sz(s) - 1;
        ur[i] = n + 1;
        while (l <= r)
        {
            int m = (l + r) / 2;
            if (a[s[m]] - d >= a[i])
            {
                ur[i] = s[m];
                l = m + 1;
            }
            else
                r = m - 1;
        }
    }

    for (int i = 1; i <= n; ++i)
    {
        ul[i]++;
        ur[i]--;
    }

    for (int i = 1; i <= n; ++i)
    {
        b.push_back(m_p(m_p(ur[i] - ul[i] + 1, i), m_p(ul[i], ur[i])));
    }
    sort(all(b));
    set<int> t;
    t.insert(0);
    t.insert(n + 1);
    for (int ii = 0; ii < n; ++ii)
    {
        int i = b[ii].fi.se;
        t.insert(i);
        auto it = t.find(i);
        --it;
        if (ur[*it] < ul[i])
            b[ii].fi.fi = 1;
        else
            b[ii].fi.fi = *it + 1;
        b[ii].fi.se = i;
        ++it;
        ++it;
        b[ii].se.fi = i;
        if (ur[i] < ul[*it])
            b[ii].se.se = n;
        else
            b[ii].se.se = *it - 1;
    }

    for (int i = 0; i < n; ++i)
    {
        ubdx(1, n, b[i].fi.fi, b[i].fi.se, b[i].se.fi, b[i].se.se, 1);
    }
}

int x;
void init(int N, std::vector<int> H)
{
    n = N;
    for (int i = 1; i <= n; ++i)
        a[i] = H[i - 1];

    x = 1;
    while (x < n && a[x] < a[x + 1])
        ++x;
    for (int i = x + 1; i <= n; ++i)
    {
        if (!(a[i] < a[i - 1]))
        {
            x = 0;
            break;
        }
    }
}

int max_towers(int L, int R, int D)
{
    ++L;
    ++R;
    if (x)
    {
        if (L < x && x < R)
        {
            if (a[L] <= a[x] - D && a[R] <= a[x] - D)
                return 2;
        }
        return 1;
    }
    if (b.empty())
    {
        d = D;
        pre();
    }

    return qryx(1, n, L, R, 1);
}

/*
7 1
10 20 30 60 50 30 10
0 4 1

7 3
10 20 60 40 50 30 70
1 5 10
2 2 100
0 6 17
*/

• Subtask 1: 4 / 4
• Subtask 2: 11 / 11
• Subtask 3: 12 / 12
• Subtask 4: 14 / 14
• Subtask 5: 0 / 17
• Subtask 6: 19 / 19
• Subtask 7: 0 / 23