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<iostream>
#include<vector>
#include<algorithm>
#define ep emplace
#define eb emplace_back
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef pair<int,int>pi;
typedef pair<ll,ll>pl;
const int inf=1e9+7;
const ll INF=1e18+7;
struct node
{
int cnt;
ll sum;
int l,r;
node(){sum=0;cnt=l=r=0;}
}tree[2222222];
int tct;
void upd(int&n,int s,int e,int x,int p)
{
tree[++tct]=tree[n];
n=tct;
tree[n].cnt++;
tree[n].sum+=p;
if(s==e)
return;
int m=s+(e-s)/2;
if(x>m)
upd(tree[n].r,m+1,e,x,p);
else
upd(tree[n].l,s,m,x,p);
return;
}
ll query(int l,int r,int k)
{
if(k<=0)
return 0;
if(k>=tree[r].cnt-tree[l].cnt)
return tree[r].sum-tree[l].sum;
if(tree[r].r==0)
return k*(tree[r].sum/tree[r].cnt);
node&ln=tree[tree[l].r];
node&rn=tree[tree[r].r];
if(k>=rn.cnt-ln.cnt)
return(rn.sum-ln.sum)+query(tree[l].l,tree[r].l,k-(rn.cnt-ln.cnt));
return query(tree[l].r,tree[r].r,k);
}
int rt[100010];
int n,st,d;
ll ans;
void dnc1(int s,int e,int as,int ae)
{
if(s>e)
return;
int m=s+(e-s)/2;
int mxi=-1;
ll cans=-1;
for(int i=as;i<=ae;i++)
{
ll cur=query(rt[i-1],rt[m],d-(st+m-i*2));
if(cur>cans)
cans=cur,mxi=i;
}
ans=max(ans,cans);
dnc1(s,m-1,as,mxi);
dnc1(m+1,e,mxi,ae);
return;
}
void dnc2(int s,int e,int as,int ae)
{
if(s>e)
return;
int m=s+(e-s)/2;
int mxi=-1;
ll cans=-1;
for(int i=as;i<=ae;i++)
{
ll cur=query(rt[m-1],rt[i],d-(i*2-st-m));
if(cur>cans)
cans=cur,mxi=i;
}
ans=max(ans,cans);
dnc2(s,m-1,as,mxi);
dnc2(m+1,e,mxi,ae);
return;
}
long long int findMaxAttraction(int n,int start,int d,int attraction[])
{
::n=n;
::st=start+1;
::d=d;
cin>>n>>st>>d;
vector<int>v(n);
for(int i=0;i<n;i++)
v[i]=attraction[i];
vector<int>cv=v;
sort(all(cv));
cv.erase(unique(all(cv)),cv.end());
int m=cv.size();
v.insert(v.begin(),0);
for(int i=0;i++<n;)
upd(rt[i]=rt[i-1],1,m,upper_bound(all(cv),v[i])-cv.begin(),v[i]);
dnc1(st,min(n,st+d-1),1,st);
dnc2(max(1,st-d+1),st,st,n);
return ans;
return 0;
}
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