# | 제출 시각UTC-0 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
1066190 | anango | 휴가 (IOI14_holiday) | C++17 | 1897 ms | 15568 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include"holiday.h"
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
#define int long long
int INF = 1LL<<60;
//method:
//consider only going right, we'll try to compute the max obtainable with d days going to the right for each d
//now, let k be the number of cities we ever visit, and thus we get to actually see attractions of d-k cities
//we obviously visit the d-k cities with maximum attraction counts
//let visit(k,d) = max sum of visits for d with visiting the first k cities and actually opening d-k of them
//let opk[d] be the optimal number of cities to visit for d days
//note that opk[d+1]>=opk[d] (proved on paper)
//thus, for each d, there's a minimum k such that visit(k,d)<=visit(k+1,d), and from now we can use k+1 instead
//so binary search on this minimum k
//ok but we can't binary search this way round since that leads to too many insertions and erasures
//so we need to do a binary search on, for each k, the minimum d such that visit(k,d)<=visit(k+1,d)
//to do this, we need a data structure that can insert values, also query the sum of the maximum t values for any t
//this can be done with a segment tree of suffix sums with lazy propagation with coordinate compression
//since we know all the attraction values in advance
//keep a PBDS of what values are currently in the set
const int sz = 262144;
vector<int> revcoords;
class SegmentTree {
public:
vector<int> tree;
int n;
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