Submission #95395

#	Time	Username	Problem	Language	Result	Execution time	Memory
95395		pranjalssh	Circle selection (APIO18_circle_selection)	C++11	45 / 100	1346 ms	54776 KiB

circle_selection

/*
	Simulate process by checking all circles in square of width 4*r. Use a kd-tree for this.
*/
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;

#define cerr cout
#define F first
#define S second
#define FOR(i,a,b) for (auto i = (a); i <= (b); ++i)
#define NFOR(i,a,b) for(auto i = (a); i >= (b); --i)
#define all(x) (x).begin(), (x).end()
#define sz(x) int(x.size())
typedef long long ll; typedef pair <int, int> ii; typedef vector <int> vi; const int inf = 1e9 + 7;
string to_string(string s) { return '"' + s + '"';}
string to_string(char s) { return string(1, s);}
string to_string(const char* s) { return to_string((string) s);}
string to_string(bool b) { return (b ? "true" : "false");}
template <typename A> string to_string(A);
template <typename A, typename B>string to_string(pair<A, B> p) {return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";}
template <typename A> string to_string(A v) {bool f = 1; string r = "{"; for (const auto &x : v) {if (!f)r += ", "; f = 0; r += to_string(x);} return r + "}";}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail> void debug_out(Head H, Tail... T) {cerr << " " << to_string(H); debug_out(T...);}
#define pr(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)


const int N = 3e5 + 10;
vi x, y, r;
struct node {
	int l, r, sum, mid;
	int aid;
	node(): l(0), r(0), sum(0), mid(0), aid(0) {}
};
vector<node> T;
vector<vector<int>> A;
vi xc, yc;
vi par;
inline bool intersect(int i, int j) {
	return (x[i] - x[j]) * 1LL * (x[i] - x[j]) + (y[i] - y[j]) * 1LL * (y[i] - y[j]) <= (r[i] + r[j]) * 1LL * (r[i] + r[j]);
}

inline int make_node() {
	T.emplace_back();
	return sz(T) - 1;
}

int build(int no, bool dim, int xl, int xr, int yl, int yr, const vi& ids) {
	if (ids.empty()) return no;
	if (no == 0) no = make_node();
	bool g = 1;
	FOR (i, 1, sz(ids) - 1) if (x[ids[i]] != x[ids[0]] or y[ids[i]] != y[ids[0]]) {g = 0; break;}
	if (g) {
		T[no].sum = sz(ids);
		A.emplace_back(ids);
		T[no].aid = sz(A)-1;
		return no;
	}
	if (dim) {
		int mid = rand() % (yr-yl) + yl;
		vi lids, rids;
		for (int it : ids) {
			if (y[it] <= yc[mid]) lids.emplace_back(it);
			else rids.emplace_back(it);
		}
		int L = build(0, 0, xl, xr, yl, mid, lids);
		int R = build(0, 0, xl, xr, mid + 1, yr, rids);
		T[no].l = L, T[no].r = R;
		T[no].sum = T[T[no].l].sum + T[T[no].r].sum;
		T[no].mid = mid;
	} else {
		int mid = rand() % (xr-xl) + xl;
		vi lids, rids;
		for (int it : ids) {
			if (x[it] <= xc[mid]) lids.emplace_back(it);
			else rids.emplace_back(it);
		}
		int L = build(0, 1, xl, mid, yl, yr, lids);
		int R = build(0, 1, mid + 1, xr, yl, yr, rids);
		T[no].l = L, T[no].r = R;
		T[no].sum = T[T[no].l].sum + T[T[no].r].sum;
		T[no].mid = mid;
	}
	return no;
}

int xu, xv, yu, yv, bc;
vi rem;

void get(int no, bool dim, int xl, int xr, int yl, int yr) {
	if (T[no].sum == 0) return;

	if (xr < xu or xl > xv) return;
	if (yr < yu or yl > yv) return;

	if (T[no].l == 0 and T[no].r == 0) {
		for (int it: A[T[no].aid]) if (intersect(it, bc)) {
			par[it] = bc;
			rem.emplace_back(it);
			T[no].sum--;
		}

		reverse(all(rem));
		vi& a = A[T[no].aid];
		int j = 0;
		FOR (i, 0, sz(a) - 1) {
			if (rem.empty() or a[i] != rem.back()) {
				a[j] = a[i];
				++j;
			} else {
				rem.pop_back();
			}
		}
		a.resize(j);
		rem.clear();
		return;
	}

	if (dim) {
		int mid = T[no].mid;
		get(T[no].l, 0, xl,xr,yl, mid);
		get(T[no].r, 0, xl, xr, mid+1, yr);
		T[no].sum = T[T[no].l].sum + T[T[no].r].sum;
	} else {
		int mid = T[no].mid;
		get(T[no].l, 1, xl, mid, yl, yr);
		get(T[no].r, 1, mid+1, xr, yl, yr);
		T[no].sum = T[T[no].l].sum + T[T[no].r].sum;
	}

}
int main()
{
	ios::sync_with_stdio(0); cin.tie(0);

	int n; cin >> n;
	x.resize(n), y.resize(n), r.resize(n);
	par = vi(n, -1);

	FOR (i, 0, n - 1) {
		cin >> x[i] >> y[i] >> r[i];
		xc.emplace_back(x[i]);
		yc.emplace_back(y[i]);
	}
	xc.emplace_back(-inf), xc.emplace_back(inf);
	yc.emplace_back(-inf), yc.emplace_back(inf);
	sort(all(xc)); xc.erase(unique(all(xc)), xc.end());
	sort(all(yc)); yc.erase(unique(all(yc)), yc.end());

	vi id(n); iota(all(id), 0);
	sort(all(id), [&](int x, int y) {
		if (r[x] == r[y]) return x < y;
		return r[x] > r[y];
	});

	T.reserve(2e7);

	T.emplace_back();
	int root = build(0, 0, 0, sz(xc)-1, 0, sz(yc)-1, id);

	for (int i : id) if (par[i] == -1) {
			bc = i;
			xu = max(x[i] - 2LL * r[i], (ll) - inf);
			xu = lower_bound(all(xc), xu) - xc.begin();
			xv = min(x[i] + 2LL * r[i], (ll)inf);
			xv = upper_bound(all(xc), xv) - xc.begin()-1;

			yu = max(y[i] - 2LL * r[i], (ll) - 1e9);
			yu = lower_bound(all(yc), yu) - yc.begin();
			yv = min(y[i] + 2LL * r[i], (ll)1e9);
			yv = upper_bound(all(yc), yv) - yc.begin()-1;

			get(root, 0, 0, sz(xc)-1, 0, sz(yc)-1);
		}

	FOR (i, 0, n - 1) cout << par[i] + 1 << " ";



	return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 0 / 12
• Subtask 3: 15 / 15
• Subtask 4: 23 / 23
• Subtask 5: 0 / 30
• Subtask 6: 0 / 13