제출 #1313835

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1313835		vlomaczk	Measures (CEOI22_measures)	C++20	0 / 100	245 ms	54144 KiB

Main

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
typedef long long ll;
using namespace __gnu_pbds;
using namespace std;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

ll base = 1<<19;
ll inf = 1e18;
vector<ll> T(2*base,0), mi(2*base,inf), ma(2*base,-inf), lazy(2*base,0); // answer, max value K[x], min value K[x]

void merge(ll v, ll l, ll r) { // Calculates max over i < j of K[j] - K[i] where K[x] = D*x + pos
	T[v] = max(T[l], T[r]);
	T[v] = max(T[v], ma[r] - mi[l]);
	ma[v] = max(ma[r], ma[l]);
	mi[v] = min(mi[r], mi[l]);
}

ll D;

void push(int v, int l, int r) {
	if(!lazy[v] || v >= base) return;
	for(auto x : {l,r}) {
		ma[x] += lazy[v];
		mi[x] += lazy[v];
		lazy[x] += lazy[v];
	}
	lazy[v] = 0;
}

void update(int v, int a, int b, int p, int k, int val, bool ustaw) {
	if(b < p || k < a) return;
	if(p<=a && b<=k) {
		if(ustaw) {
			mi[v] = ma[v] = val;
			T[v] = lazy[v] = 0;
		} else {
			mi[v] += val;
			ma[v] += val;
			lazy[v] += val;
		}
		return;
	}
	int l =2*v; int r=l+1; int mid = (a+b)/2;
	push(v,l,r);
	update(l,a,mid,p,k,val,ustaw);
	update(r,mid+1,b,p,k,val,ustaw);
	merge(v,l,r);
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);

	ll n, m;
	cin >> n >> m >> D;
	vector<ll> x(n), y(m);
	for(ll i=0; i<n; ++i) cin >> x[i];
	for(ll i=0; i<m; ++i) cin >> y[i];
	vector<pair<ll, ll>> pary;
	for(ll i=0; i<n; ++i) {
		pary.push_back({x[i], i});
	}
	for(ll i=0; i<m; ++i) {
		pary.push_back({y[i], i+n});
	}
	sort(pary.begin(), pary.end());
	ll cnt = 0;
	vector<ll> idx(n+m);
	ordered_set<ll> os;
	for(auto[a,b] : pary) {
		idx[b] = cnt++;
	}	
	for(ll i=0; i<n; ++i) {
		os.insert(idx[i]);
		update(1,0,base-1, idx[i], idx[i], D*os.order_of_key(idx[i]) - x[i], 1);
		update(1,0,base-1, idx[i]+1, base-1, D, 0);
	}
	for(ll i=0; i<m; ++i) {
		os.insert(idx[i+n]);
		update(1,0,base-1, idx[i+n], idx[i+n], D*os.order_of_key(idx[i+n]) - y[i], 1);
		update(1,0,base-1, idx[i+n]+1, base-1, D, 0);
		cout << T[1]/2;
		if(T[1]%2==1) cout << ".5";
		cout << " ";
	}
	cout << "\n";

	return 0;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 14
• Subtask 3: 0 / 35
• Subtask 4: 0 / 41