제출 #420259

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
420259		amoo_safar	Event Hopping 2 (JOI21_event2)	C++17	100 / 100	301 ms	25508 KiB

event2

// vaziat meshki-ghermeze !
#include <bits/stdc++.h>

#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x << " : " << x << '\n'

using namespace std;

typedef long long ll;
typedef long double ld;
typedef string str;
typedef pair<ll, ll> pll;

const ll Mod = 1000000007LL;
const int N = 2e5 + 10;
const ll Inf = 2242545357980376863LL;
const int Log = 20;

int nx[Log][N];
int mn[N];
int l[N], r[N];

int Solve(int lq, int rq){
	int ans = 0;
	for(int lg = Log - 1; lg >= 0; lg--){
		if(nx[lg][lq] <= rq){
			lq = nx[lg][lq];
			ans += (1 << lg);
		}
	}
	return ans;
}

int main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int n, k;
	cin >> n >> k;
	vector<int> V;
	for(int i = 1; i <= n; i++){
		cin >> l[i] >> r[i];
		V.pb(l[i]); V.pb(r[i]);
	}
	sort(all(V));
	V.resize(unique(all(V)) - V.begin());
	int m = V.size();
	fill(mn, mn + m + 1, m);
	for(int i = 1; i <= n; i++){
		l[i] = lower_bound(all(V), l[i]) - V.begin();
		r[i] = lower_bound(all(V), r[i]) - V.begin();
		mn[l[i]] = min(mn[l[i]], r[i]);
	}

	for(int i = m - 1; i >= 0; i--)
		mn[i] = min(mn[i], mn[i + 1]);
	for(int i = 0; i <= m; i++) nx[0][i] = mn[i];
	for(int lg = 0; lg + 1 < Log; lg++)
		for(int i = 0; i <= m; i++)
			nx[lg + 1][i] = nx[lg][ nx[lg][i] ];

	if(Solve(0, m - 1) < k)
		return cout << "-1\n", 0;
	int sm = Solve(0, m - 1);
	set< pair<int, int> > st;
	st.insert({0, m - 1});
	
	vector<int> pk;
	for(int i = 1; i <= n; i++){
		auto it = st.upper_bound({l[i], Inf});
		if(it == st.begin())
			continue;
		it --;
		auto seg = *it;
		if(seg.S < r[i])
			continue;

		int ft = sm - Solve(seg.F, seg.S) + Solve(seg.F, l[i]) + Solve(r[i], seg.S) + 1;
		if(ft < k)
			continue;
		st.erase(seg);
		if(seg.F != l[i])
			st.insert({seg.F, l[i]});
		if(seg.S != r[i])
			st.insert({r[i], seg.S});
		sm = ft;
		pk.pb(i);
	}
	for(int i = 0; i < k; i++) cout << pk[i] << '\n';
	return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 1 / 1
• Subtask 3: 31 / 31
• Subtask 4: 61 / 61