제출 #847398

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
847398		Lib	봉쇄 시간 (IOI23_closing)	C++17	0 / 100	64 ms	13512 KiB

closing

#include "closing.h"
#include <bits/stdc++.h>
using namespace std;
long long pos[200003];
long long cost1[200003];
long long cost2[200003];
long long pfx1[200003];
long long pfx2[200003];
long long maxroll[200003]; //maxroll[i]=a <==> de "tra no" 1 khoang toi da =a, can rollback it nhat i vi tri tu loai 1 xuong loai 0
long long minupd[200003]; //minupt[i]=k <==> voi 1 budget<=i; upgrade duoc toi da k vi tri tu loai 1 len loai 2
struct cus{
bool operator()(const long long &x, const long long &y){
	return x>y;
}
};
int max_score(int N, int X, int Y, long long K, std::vector<int> U, std::vector<int> V, std::vector<int> W)
{
	int n=N;
	int x=X;
	int y=Y;
	long long k=K;
	for(int i=0;i<n-1;i++){
		cin>>U[i]>>V[i]>>W[i];
		pos[i+1]=pos[i]+W[i];
	}
	for(int i=0;i<n;i++){
		cost1[i]=min(abs(pos[i]-pos[x]),abs(pos[i]-pos[y]));
		cost2[i]=max(abs(pos[i]-pos[x]),abs(pos[i]-pos[y]))-cost1[i];
		pfx1[i+1]=pfx1[i]+cost1[i];
	}
	priority_queue <long long> pq;
	long long mcost=k;
	//case 1: chi nang cap cac dinh len toi da 1
	for(int i=0;i<n;i++){
		pq.push(-cost1[i]);
	}
	long long ans=0;
	while(mcost>0&&!pq.empty()){
		mcost+=pq.top();
		ans++;
		pq.pop();
	}
	if(mcost<0){
		ans--;
	}
	long long tans=0;
	tans=max(tans,ans);
	while(!pq.empty()){
		pq.pop();
	}
	///case 2: ton tai dinh duoc nang cap len 2 => tat ca cac diem trong doan (l,r) deu da duoc nang cap len loai 1. sau do tham nhu tren
	mcost=k-(pfx1[y+1]-pfx1[x]);
	if(mcost>0){
		ans=y-x+1;
		for(int i=0;i<n;i++){
			pq.push(-cost2[i]);
		}
		for(int i=0;i<n;i++){
			if(i<x||i>y){
				pq.push(-cost1[i]);
			}
		}
		while(!pq.empty()&&mcost>0){
			ans++;
			mcost+=pq.top();
			pq.pop();
		}
		if(mcost<0){
			ans--;
		}
		tans=max(tans,ans);
	}
	return tans;
}

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