Submission #870131

#	Time	Username	Problem	Language	Result	Execution time	Memory
870131		NK_	Road Construction (JOI21_road_construction)	C++17	100 / 100	4267 ms	27184 KiB

road_construction

// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
 
using namespace std;
 
#define nl '\n'
#define pb push_back
#define pf push_front
 
#define mp make_pair
#define f first
#define s second
#define sz(x) int(x.size())
 
template<class T> using V = vector<T>;
using pi = pair<int, int>;
using vi = V<int>;
using vpi = V<pi>;
 
using ll = long long;
using pl = pair<ll, ll>;
using vpl = V<pl>;
using vl = V<ll>;
 
using db = long double;
using str = string;
 
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;

template<class T> using pq = priority_queue<T, V<T>, greater<T>>;
template<class T> using iset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; 
 
const int MOD = 1e9 + 7;
const ll INFL = ll(1e17);
const int LG = 19;

int main() {
	cin.tie(0)->sync_with_stdio(0);

	// rotate 45 (x, y) -> (x + y, x - y)
	// Change Manhattan Distance to C. Distance (max(∆x, ∆y))

	int N, K; cin >> N >> K;
	vpl A(N); for(auto& x : A) cin >> x.f >> x.s;

	for(auto& x : A) x = mp(x.f + x.s, x.f - x.s);
	sort(begin(A), end(A));

	V<array<ll, 3>> R; 
	for(int i = 0; i < N; i++) {
		int j = i;
		while(j < N && A[i].f == A[j].f) j++;
		R.pb(array<ll, 3>{A[i].f, i, j - 1});
		i = j - 1;
	}

	int M = sz(R);

	auto dist = [&](const pl &a, const pl &b) -> ll { return max(abs(a.f - b.f), abs(a.s - b.s)); };

	iset<pl> B; vl ANS;

	auto get = [&](ll d, int t) {
		// d -> distance within
		// t -> get pairs

		B = {}; ll ans = 0; vl res;
		for(int l = 0, r = 0; r < M; r++) {
			// REM l(s)
			while(l <= r && (R[r][0] - R[l][0]) > d) {
				for(int i = R[l][1]; i <= R[l][2]; i++) B.erase(mp(A[i].s, i));
				l++;
			}

			// ADD r + COUNT PAIRS (for people in r);
			for(int i = R[r][1]; i <= R[r][2]; i++) {
				B.insert(mp(A[i].s, i));
				int li = B.order_of_key(mp(A[i].s - d, -MOD));
				int ri = B.order_of_key(mp(A[i].s + d, MOD)) - 1;

				ans += ri - li;
				if (t) for(int x = li; x <= ri; x++) {
					int j = (*B.find_by_order(x)).s;
					if (i == j) continue;

					res.pb(dist(A[i], A[j]));
				}
			}
		}	

		if (t) ANS = res;

		return ans;
	};


	ll lo = 0, hi = 5LL * MOD;
	while(lo < hi) {
		ll mid = (lo + hi) / 2;
		if (get(mid, 0) > K) hi = mid;
		else lo = mid + 1;
	}

	get(lo - 1, 1); sort(begin(ANS), end(ANS));
	while(sz(ANS) < K) ANS.pb(lo);

	for(auto& x : ANS) cout << x << nl;

	exit(0-0);
} 	 

• Subtask 1: 5 / 5
• Subtask 2: 6 / 6
• Subtask 3: 7 / 7
• Subtask 4: 20 / 20
• Subtask 5: 27 / 27
• Subtask 6: 35 / 35