Submission #768582

#	Time	Username	Problem	Language	Result	Execution time	Memory
768582		jmyszka2007	Financial Report (JOI21_financial)	C++17	100 / 100	182 ms	15348 KiB

Main

#include <bits/stdc++.h>
using namespace std;
#define st first
#define nd second
#define pi pair<int, int>
#define pb push_back
constexpr int LIM = 300'000;
constexpr int base = (1 << 19);
int dp[LIM + 10];
pi tab[LIM + 10];
int ran[LIM + 10];
int tri[2 * base];
int tri2[2 * base];
void upd(int v, int x) {
	v += base;
	while(v > 0) {
		tri[v] = min(tri[v], x);
		v /= 2;
	}
}
int que(int l, int r) {
	l += base;
	r += base;
	int ans = 1e9;
	while(l <= r) {
		if(l & 1) {
			ans = min(ans, tri[l]);
		}
		if(!(r & 1)) {
			ans = min(ans, tri[r]);
		}
		l = (l + 1) / 2;
		r = (r - 1) / 2;
	}
	return ans;
}
void upd2(int l, int r, int x) {
	l += base;
	r += base;
	while(l <= r) {
		if(l & 1) {
			tri2[l] = max(tri2[l], x);
		}
		if(!(r & 1)) {
			tri2[r] = max(tri2[r], x);
		}
		l = (l + 1) / 2;
		r = (r - 1) / 2;
	}
}
int que2(int v) {
	v += base;
	int ans = 0;
	while(v > 0) {
		ans = max(ans, tri2[v]);
		v /= 2;
	}
	return ans;
}
int main() {
	int n, d;
	cin >> n >> d;
	for(int i = 0; i < 2 * base; i++) {
		tri[i] = 1e9;
	}
	for(int i = 1; i <= n; i++) {
		cin >> tab[i].st;
		tab[i].nd = -i;
		upd(i, tab[i].st);
	}
	int ost = tab[n].st;
	vector<pi>prz;
	for(int i = n - d; i >= 1; i--) {
		if(!prz.size() || prz[0].nd <= tab[i].st) {
			ran[i] = n;
		}
		else {
			int l = 0;
			int r = prz.size() - 1;
			while(l < r) {
				int mid = (l + r + 1) / 2;
				if(prz[mid].nd <= tab[i].st) {
					r = mid - 1;
				}
				else {
					l = mid;
				}
			}
			ran[i] = prz[l].st + d - 1;
		}
		int mn = que(i, i + d - 1);
		while(prz.size() && prz.back().nd <= mn) {
			prz.pop_back();
		}
		prz.pb({i, mn});
	}
	for(int i = n; i >= n - d + 1; i--) {
		ran[i] = n;
	}
	sort(tab + 1, tab + n + 1);
	int ans = 0;
	for(int i = 1; i <= n; i++) {
		tab[i].nd *= -1;
		dp[tab[i].nd] = que2(tab[i].nd) + 1;
		upd2(tab[i].nd, ran[tab[i].nd], dp[tab[i].nd]);
		if(tab[i].st >= ost && ran[tab[i].nd] == n) {
			ans = max(ans, dp[tab[i].nd]);
		}
	}
	cout << ans << '\n';
}

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 20 / 20
• Subtask 4: 12 / 12
• Subtask 5: 5 / 5
• Subtask 6: 35 / 35