제출 #1144863

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1144863		monkey133	Bouquet (EGOI24_bouquet)	C++20	100 / 100	86 ms	16056 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll l[200005], r[200005];

vector<ll> fenwick(200005,0);

ll rmq(ll i)
{
	if (i==0) return 0;
	ll maxx=0;
	while (i>0)
	{
		maxx=max(maxx, fenwick[i]);
		i-=i&(-i);
	}
	return maxx;
}
void update(ll i, ll v)
{
	while (i<fenwick.size())
	{
		fenwick[i] = max(fenwick[i], v);
		i+=i&(-i);
	}
}

vector<pi> upd[200005];


int main()
{
	ll n;
	cin >> n;
	for (ll i=1; i<=n; ++i) cin >> l[i] >> r[i];
	ll ans = 0;
	for (ll i=1; i<=n; ++i)
	{
		for (pi u: upd[i]) update(u.f, u.s);
		ll dp = rmq(max(0LL, i - l[i]-1)) + 1;
		ans = max(dp, ans);
		upd[min(n + 1, i + r[i] + 1)].pb(mp(i, dp));
	}
	cout << ans << endl;
	return 0;
}

• Subtask 1: 8 / 8
• Subtask 2: 16 / 16
• Subtask 3: 28 / 28
• Subtask 4: 18 / 18
• Subtask 5: 30 / 30