제출 #623077

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
623077		CSQ31	Road Construction (JOI21_road_construction)	C++17	0 / 100	3591 ms	225764 KiB

road_construction

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
//consider the case where we only need k smallest from two different set of points
//if can do this in some reasonable time then apply zs bit trick
//+log factor
//need <= log^3 maybe?
//ok two set need log^2(?)
#define all(a) a.begin(),a.end()
#define pb push_back
#define owo ios_base::sync_with_stdio(0);cin.tie(0);
const int MAXN = 2000000;
vector<ll>glo;
multiset<int>fen[MAXN];
multiset<int>::iterator it[MAXN];
vector<int>touch;
void upd(int i,int n,int v){
	for(;i<=n;i+=i&(-i))fen[i].insert(v);
}
void query(int i){
	while(i>0){
		touch.pb(i);
		i-=i&(-i);
	}
}
#define fi first
#define se second 
typedef pair<int,int> pii;
int main()
{
	owo
	int n,k;
	cin>>n>>k;
	vector<array<int,2>>p(n);
	vector<int>cy(n);
	for(int i=0;i<n;i++)cin>>p[i][0]>>p[i][1];
	sort(all(p));
	vector<int>crd;
	for(int i=0;i<n;i++)crd.push_back(p[i][1]);
	sort(all(crd));
	crd.resize(unique(all(crd)) - crd.begin());
	for(int i=0;i<n;i++)cy[i] = lower_bound(all(crd),p[i][1]) - crd.begin()+1;
	int m = crd.size();
	if(k==1){
		ll ans = 1e18;
		for(int i=0;i<n;i++){
			touch.clear();
			query(cy[i]);
			int s = touch.size();
			for(int j=0;j<s;j++){
				if(fen[touch[j]].empty())continue;
				ll c = p[i][0] + p[i][1];
				c-=*fen[touch[j]].begin();
				ans = min(ans,c);
			}
			upd(cy[i],m,p[i][0]+p[i][1]);
		}
		//cout<<ans<<'\n';
		//y(i) < y(j) case
		//we use x(i) - y(i) - (x(j)-y(j)) 
		for(int i=1;i<=m;i++)fen[i].clear(); //i need to do the opposite directly
		for(int i=0;i<n;i++){
			touch.clear();
			cy[i] = m-cy[i]+1;
			query(cy[i]-1);
			int s = touch.size();
			for(int j=0;j<s;j++){
				if(fen[touch[j]].empty())continue;
				ll c = p[i][0] - p[i][1];
				c-=*fen[touch[j]].begin();
				ans = min(ans,c);
			}
			upd(cy[i],m,p[i][0]-p[i][1]);
		}
		cout<<ans<<'\n';
	}
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 6
• Subtask 3: 0 / 7
• Subtask 4: 0 / 20
• Subtask 5: 0 / 27
• Subtask 6: 0 / 35