제출 #756443

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
756443		Red_Inside	송신탑 (IOI22_towers)	C++17	23 / 100	4093 ms	5204 KiB

towers

#include "towers.h"
#include <bits/stdc++.h>

//#pragma GCC optimize("O1,O2,O3,Ofast,unroll-loops")

#define ll long long
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define o cout<<"BUG"<<endl;
#define FOR(i, j, n) for(int j = i; j < n; ++j)
#define forn(i, j, n) for(int j = i; j <= n; ++j)
#define nfor(i, j, n) for(int j = n; j >= i; --j)
#define all(v) v.begin(), v.end()
#define ld long double
#define ull unsigned long long
 
using namespace std;
const int maxn=1e5+10,LOG=17,mod=1e9+7;
int block = 650, timer = 0;
const double eps = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 
#define bt(i) (1 << (i))
//#define int ll
const long long inf=2e18;
#define y1 yy
#define pii pair <int, int>

int n, a[maxn], t[4*maxn];

void build(int v, int tl, int tr)
{
	if(tl == tr)
	{
		t[v] = a[tl];
		return;
	}
	build(v * 2, tl, (tl + tr) / 2);
	build(v * 2 + 1, (tl + tr) / 2 + 1, tr);
	t[v] = max(t[v * 2], t[v * 2 + 1]);
}

int getmax(int v, int tl, int tr, int l, int r)
{
	if(l <= tl && tr <= r) return t[v];
	if(l > tr || r < tl) return 0;
	return max(getmax(v * 2, tl, (tl + tr) / 2, l, r), 
	getmax(v * 2 + 1, (tl + tr) / 2 + 1, tr, l, r));
}

void init(int N, std::vector<int> H) {
	n = N;
	forn(1, i, n)
	{
		a[i] = H[i-1];
	}
	build(1, 1, n);
}

int Ans;

int max_towers(int l, int r, int d) {
	l++;
	r++;
	if(l == 1 && r == n && Ans != 0) return Ans;
	vector <pii> vec;
	forn(l, i, r)
	{
		vec.pb({a[i], i});
	}
	sort(all(vec));
	set <int> st;
	st.insert(n + 1);
	st.insert(0);
	for(auto it : vec)
	{
		int pos = it.s;
		auto iter = st.lower_bound(pos);
		int r = *iter;
		iter--;
		int l = *iter;
		if((l == 0 || getmax(1, 1, n, l + 1, pos-1) >= a[pos]+d) && 
		(r == n+1 || getmax(1, 1, n, pos+1, r-1) >= a[pos]+d))
		{
			st.insert(pos);
		}
	}
	if(l == 1 && r == n) Ans = (int)st.size() - 2;
	return (int)st.size() - 2;
}

• Subtask 1: 0 / 4
• Subtask 2: 11 / 11
• Subtask 3: 12 / 12
• Subtask 4: 0 / 14
• Subtask 5: 0 / 17
• Subtask 6: 0 / 19
• Subtask 7: 0 / 23