【tmp】决策单调性dp的一种类似整体二分的写法

portal-->

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN=5*(1e5)+10;
int a[MAXN],num[MAXN];
double Sqrt[MAXN],ans[MAXN];
int n,m;
void solve(int l,int r,int ansl,int ansr);
double get_ans(int i,int j);
int Abs(int x){return x<0?-x:x;}

int main(){
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
#endif
	scanf("%d",&n);
	for (int i=1;i<=n;++i) Sqrt[i]=(double)sqrt(i);
	for (int i=1;i<=n;++i) scanf("%d",a+i),num[i]=i;
	solve(1,n,1,n);
	for (int i=1;i<=n/2;++i)
		swap(a[i],a[n-i+1]),swap(num[i],num[n-i+1]);
	solve(1,n,1,n);
	for (int i=1;i<=n;++i) 
		printf("%d\n",(int)ceil(ans[i]-a[n-i+1]));
}

void solve(int l,int r,int ansl,int ansr){
	if (l>r) return;
	int mid=l+r>>1,pos=0;
	double mx=0;
	for (int i=ansl;i<=ansr&&i<=mid;++i)
		if (get_ans(mid,i)>mx) 
			mx=get_ans(mid,i),pos=i;
	ans[num[mid]]=max(ans[num[mid]],mx);
	solve(l,mid-1,ansl,pos);
	solve(mid+1,r,pos,ansr);
}

double get_ans(int i,int j){
	return a[j]+Sqrt[Abs(i-j)];
}