P3515 [POI2011] Lightning Conductor (decision monotony of divide and conquer)

P3515 [POI2011]Lightning Conductor

May be converted to the formula: $ p> = a_j-a_i + sqrt (ij) (j <i) $

Where $ j> i $, and the formula can be inverted to give

Below a picture to prove it is to meet the decision-making monotony

The $ a_j + sqrt (ij) $ represented on a coordinate system

Apparently $ sqrt (ij) $ growth has slowed down

Curve A $ $ $ b $ after the curve is not more than the stand

Both directions monotonic points decision rule, they can take $ max $

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int read(){
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') f=f&&(c!='-'),c=getchar();
    while('0'<=c&&c<='9') x=x*10+c-48,c=getchar();
    return f?x:-x;
}
#define N 500005
int n,a[N]; double p1[N],p2[N],w;
void solve1(int l,int r,int dl,int dr){
    int m=(l+r)/2,dm=dl;
    for(int i=dl;i<=m&&i<=dr;++i)
        if(p1[m]<(w=a[i]-a[m]+sqrt(m-i)))
            p1[m]=w,dm=i;
    if(l<m) solve1(l,m-1,dl,dm);
    if(m<r) solve1(m+1,r,dm,dr);
}
void solve2(int l,int r,int dl,int dr){
    int m=(l+r)/2,dm=dr;
    for(int i=dr;i>=m&&i>=dl;--i)
        if(p2[m]<(w=a[i]-a[m]+sqrt(i-m)))
            p2[m]=w,dm=i;
    if(l<m) solve2(l,m-1,dl,dm);
    if(m<r) solve2(m+1,r,dm,dr);
}
int main(){
    n=read();
    for(int i=1;i<=n;++i) a[i]=read();
    solve1(1,n,1,n);
    solve2(1,n,1,n);
    for(int i=1;i<=n;++i)
        printf("%d\n",(int)ceil(max(p1[i],p2[i])));
}