제출 #779028

#제출 시각아이디문제언어결과실행 시간메모리
779028OrazBJust Long Neckties (JOI20_ho_t1)C++14
100 / 100
254 ms18576 KiB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,
             tree_order_statistics_node_update>
    ordered_set;
//Dijkstra->set
//set.find_by_order(x) x-position value
//set.order_of_key(x) number of strictly less elements don't need *set.??
#define N 100005
#define wr cout << "Continue debugging\n";
#define all(x) (x).begin(), (x).end()
#define ll long long int
#define pii pair <int, int>
#define pb push_back
#define ff first
#define ss second

map <int,int> cur;

int main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	vector<int> a(n+2), b(n+1), old_a(n+2);
	for (int i = 1; i <= n+1; i++) cin >> a[i], cur[a[i]] = 1e9+7, old_a[i] = a[i];
	for (int i = 1; i <= n; i++) cin >> b[i];
	sort(all(a)); sort(all(b));
	vector<int> suff(n+3, 0);
	for (int i = n+1; i >= 1; i--){
		suff[i] = max(suff[i+1], a[i]-b[i-1]);
	}
	int mx = 0;
	for (int i = 1; i <= n+1; i++){
		cur[a[i]] = min(cur[a[i]], max(mx, suff[i+1]));
		mx = max(mx, a[i]-b[i]);
	}
	for (int i = 1; i <= n+1; i++) cout << cur[old_a[i]] << ' ';
}	
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...