Submission #58825

#	Time	Username	Problem	Language	Result	Execution time	Memory
58825		Benq	Segments (IZhO18_segments)	C++14	22 / 100	1893 ms	41984 KiB

segments

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 200001;

int n,t,ans, cbucket[MX], L[MX], R[MX];
vi buckety[400], bucketyx[400];
vpi lst[400];

bool cmp(int l, int r) {
    if (L[l] != L[r]) return L[l] < L[r];
    return l < r;
}

void rebuild() {
    vpi v;
    F0R(i,400) {
        v.insert(v.end(),all(lst[i]));
        buckety[i].clear(), bucketyx[i].clear(), lst[i].clear();
    }

    F0R(i,400) {
        FOR(j,i*sz(v)/400,(i+1)*sz(v)/400) {
            int id = v[j].s;
            cbucket[id] = i;
            buckety[i].pb(R[id]);
            bucketyx[i].pb(R[id]-L[id]);
            lst[i].pb(v[j]);
        }

        sort(all(buckety[i]));
        sort(all(bucketyx[i]));
    }
}

template<class T> void ins(vector<T>& v, T x, int ad = 1) {
    auto it = lb(all(v),x);
    if (ad == 1) v.insert(it,x);
    else {
        assert(it != v.end());
        v.erase(it);
    }
}

void del(int id) {
    int b = cbucket[id];
    ins(buckety[b],R[id],-1);
    ins(bucketyx[b],R[id]-L[id],-1);
    ins(lst[b],{L[id],id},-1);
}

void upd(int id, int l, int r) {
    L[id] = l, R[id] = r;
    int b = 0; 
    while (b < 399 && (sz(lst[b]) == 0 || lst[b].back() < mp(L[id],id))) b ++;

    cbucket[id] = b;
    ins(buckety[b],R[id]);
    ins(bucketyx[b],R[id]-L[id]);
    ins(lst[b],{L[id],id});
}

int hix, loy, loyx;

int manual(int x, int l) {
    int t = 0;
    for (pi i: lst[x]) {
        if (i.f > hix) continue;
        if (i.f > l) t += ((R[i.s]-L[i.s]) >= loyx);
        else t += (R[i.s] >= loy);
    }
    return t;
}

int atLeast(vi& x, int y) {
    return distance(lb(all(x),y),x.end());
}

int query(int l, int r, int k) {
    if (k > r-l+1) return 0;
    hix = r-k+1, loy = l+k-1, loyx = k-1;
    // y >= loy, y >= x+loyx 
    
    int ans = 0;
    int clst = -MOD;
    F0R(i,400) if (sz(lst[i])) {
        if (lst[i].back().f > hix) {
            ans += manual(i,l);
            break;
        } else if (lst[i].back().f > l) {
            if (clst <= l) ans += manual(i,l);
            else ans += atLeast(bucketyx[i],loyx);
        } else {
            ans += atLeast(buckety[i],loy);
        }
        clst = lst[i].back().f;
    }   

    return ans;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> t;
    
    int cur = 0, nex = 1;
    F0R(i,n) {
        int z; cin >> z;
        if (z == 1) {
            int l,r; cin >> l >> r;
            l ^= (t*ans), r ^= (t*ans);
            if (l > r) swap(l,r);
            upd(nex++,l,r);
        } else if (z == 2) {
            int id; cin >> id;
            del(id);
        } else {
            int l,r,k; cin >> l >> r >> k;
            l ^= (t*ans), r ^= (t*ans);
            if (l > r) swap(l,r);
            ans = query(l,r,k);
            cout << ans << "\n";
        }
        cur ++;
        if (cur % 500 == 0) rebuild();
    }
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 7 / 7
• Subtask 2: 15 / 15
• Subtask 3: 0 / 16
• Subtask 4: 0 / 17
• Subtask 5: 0 / 20
• Subtask 6: 0 / 25