제출 #512250

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
512250		fatemetmhr	Port Facility (JOI17_port_facility)	C++17	78 / 100	6093 ms	840784 KiB

port_facility

//  ~Be name khoda~  //

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define pb       push_back
#define mp       make_pair
#define all(x)   x.begin(), x.end()
#define fi       first
#define se       second
#define cl       clear
#define endll    '\n'

const int maxn  =  1e6   + 10;
const int maxn5 =  1e6   + 10;
const int maxnt =  8e6   + 10;
const int maxn3 =  1e3   + 10;
const int mod   =  1e9   +  7;
const ll  inf   =  2e18;

vector <int> adj, have, adj2[maxn5];
int ty[maxn5], a[maxn5], b[maxn5];
int per1[maxn5], per2[maxn5];
int per1c[maxn5], per2c[maxn5];
bool mark[maxn5];
vector <int> seg1[maxnt], seg2[maxnt];
vector <pair<int, int>> av;
set <pair<int, int>> seg;

int ms, n;

inline bool cmp1(int i, int j){return b[i] < b[j];}
inline bool cmp2(int i, int j){return a[i] > a[j];}
inline bool cmp3(int i, int j){return a[i] < a[j];}
inline bool cmp4(int i, int j){return b[i] < b[j];}

inline void build1(int l, int r, int v){
	int p = lower_bound(per1c, per1c + n, l) - per1c;
	for(int i = p; i < n && a[per1[i]] < r; i++)
		seg1[v].pb(per1[i]);
	sort(all(seg1[v]), cmp1);
	if(r - l == 1)
		return;
	int mid = (l + r) >> 1;
	build1(l, mid, v * 2);
	build1(mid, r, v * 2 + 1);
	return;
}

inline void build2(int l, int r, int v){
	int p = lower_bound(per2c, per2c + n, l) - per2c;
	for(int i = p; i < n && b[per2[i]] < r; i++)
		seg2[v].pb(per2[i]);
	sort(all(seg2[v]), cmp2);
	if(r - l == 1)
		return;
	int mid = (l + r) >> 1;
	build2(l, mid, v * 2);
	build2(mid, r, v * 2 + 1);
	return;
}

inline void get1(int l, int r, int lq, int rq, int val, int v){
	if(rq <= l || r <= lq)
		return;
	if(lq <= l && r <= rq){
		while(!seg1[v].empty() && b[seg1[v].back()] >= val){
			adj.pb(seg1[v].back());
			//cout << "aha " << seg1[v].back() << ' ' << val << ' ' << b[seg1[v].back()] << endl;
			seg1[v].pop_back();
		}
		return;
	}
	int mid = (l + r) >> 1;
	get1(l, mid, lq, rq, val, v * 2);
	get1(mid, r, lq, rq, val, v * 2 + 1);
	return;
}

inline void get2(int l, int r, int lq, int rq, int val, int v){
	if(rq <= l || r <= lq)
		return;
	if(lq <= l && r <= rq){
		while(!seg2[v].empty() && a[seg2[v].back()] <= val){
			adj.pb(seg2[v].back());
			seg2[v].pop_back();
		}
		return;
	}
	int mid = (l + r) >> 1;
	get2(l, mid, lq, rq, val, v * 2);
	get2(mid, r, lq, rq, val, v * 2 + 1);
	return;
}


inline void dfs2(int v){
	mark[v] = true;
	adj.clear();
	vector <int> have;
	get1(0, ms, a[v], b[v] + 1, b[v], 1); // laghal
	get2(0, ms, a[v], b[v] + 1, a[v], 1); // hadeaksar
	for(auto u : adj)
		adj2[v].pb(u);
	for(auto u : adj2[v]){
		//cout << "e " << v << ' ' << u << endl;
		if(!mark[u])
			dfs2(u);
	}
	return;
}

inline void dfs(int v){
	mark[v] = true;
	for(auto u : adj2[v]) if(!mark[u]){
		ty[u] = (ty[v] ^ 1);
		dfs(u);
	}
	return;
}

inline void check(){
	av.clear();
	for(auto i : have){
		mark[i] = false;
		av.pb({a[i], i});
		av.pb({b[i], i});
	}
	sort(all(av));
	seg.clear();
	for(auto [u, i] : av){
		if(u == a[i]){
			seg.insert({a[i], i});
		}
		else{
			seg.erase({a[i], i});
			auto it = seg.lower_bound(mp(a[i], 0));
			if(it != seg.end()){
				cout << 0 << endl;
				exit(0);
			}
		}
	}
	return;
}

int main()
{
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	
	cin >> n;
	
	ms = 2 * n + 1;

	for(int i = 0; i < n; i++){
		cin >> a[i] >> b[i];
		a[i]--; b[i]--;
		per1[i] = i;
		per2[i] = i;
	}
	sort(per1, per1 + n, cmp3);
	sort(per2, per2 + n, cmp4);
	for(int i = 0; i < n; i++){
		per1c[i] = a[per1[i]];
		per2c[i] = b[per2[i]];
	}
	
	build1(0, ms, 1);
	build2(0, ms, 1);
	
	for(int i = 0; i < n; i++) if(!mark[i]){
		dfs2(i);
	}
	
	fill(mark, mark + n + 10, false);
	
	int cnt = 0;
	for(int i = 0; i < n; i++) if(!mark[i]){
		ty[i] = 0;
		dfs(i);
		cnt++;
	}
	for(int i = 0; i < n; i++) if(ty[i] == 0){
		have.pb(i);
	}
	check();
	have.clear();
	for(int i = 0; i < n; i++) if(ty[i] == 1)
		have.pb(i);
	check();
	ll ans = 1;
	while(cnt--){
		ans *= 2;
		ans %= mod;
	}
	cout << ans << endl;
}

• Subtask 1: 10 / 10
• Subtask 2: 12 / 12
• Subtask 3: 56 / 56
• Subtask 4: 0 / 22