[POI2011]Lightning Conductor

题目描述

Progressive climate change has forced the Byteburg authorities to build a huge lightning conductor that would protect all the buildings within the city.

These buildings form a row along a single street, and are numbered from  to .

The heights of the buildings and the lightning conductor are non-negative integers.

Byteburg's limited funds allow construction of only a single lightning conductor.

Moreover, as you would expect, the higher it will be, the more expensive.

The lightning conductor of height  located on the roof of the building  (of height ) protects the building  (of height ) if the following inequality holds:

 where  denotes the absolute value of the difference between  and .

Byteasar, the mayor of Byteburg, asks your help.

Write a program that, for every building , determines the minimum height of a lightning conductor that would protect all the buildings if it were put on top of the building .

输入格式

In the first line of the standard input there is a single integer  () that denotes the number of buildings in Byteburg.

Each of the following  lines holds a single integer  () that denotes the height of the -th building.

输出格式

Your program should print out exactly  lines to the standard output.

The -th line should give a non-negative integer  denoting the minimum height of the lightning conductor on the -th building.

题意翻译

给定一个长度为 nn 的序列 {a_n}{an​}，对于每个 iin [1,n]i∈[1,n] ，求出一个最小的非负整数 pp ，使得 forall jin[1,n]∀j∈[1,n]，都有 a_jle a_i+p-sqrt{|i-j|}aj​≤ai​+p−∣i−j∣​

1 le n le 5 imes 10^{5}1≤n≤5×105，0 le a_i le 10^{9}0≤ai​≤109 。

输入输出样例

输入 #1复制

6
5
3
2
4
2
4

输出 #1复制

2
3
5
3
5
4

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;
int q[500005],R[500005],a[500005],n,b[500005];
double sq[500005],f1[500005],f2[500005];
double getValue(int x,int y)
{
    return a[y]+(sq[x-y]);
}
int binary(int x,int y)
{
    int l=x,r=n+1;
    while (l<r)
    {
        int mid=(l+r)/2;
        if (getValue(mid,y)<=getValue(mid,x)) r=mid;
        else l=mid+1;
    }
    return l;
}
int main()
{int i,j;
    scanf("%d",&n);
    for (i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        sq[i]=sqrt((double)i);
    }
    int h=1,t=0;
    for (i=1;i<=n;i++)
    {
        while (h<t&&R[t-1]>=binary(i,q[t])) t--;
        R[t]=binary(i,q[t]);
        q[++t]=i;
        while (h<t&&R[h]<=i) h++;
        f1[i]=getValue(i,q[h]);
    }
    h=1,t=0;
    for (i=1;i<=n;i++)
    b[i]=a[n-i+1];
    for (i=1;i<=n;i++)
    a[i]=b[i];
    for (i=1;i<=n;i++)
    {    
        while (h<t&&R[t-1]>=binary(i,q[t])) t--;
        R[t]=binary(i,q[t]);
        q[++t]=i;
        while (h<t&&R[h]<=i) h++;
        f2[n-i+1]=getValue(i,q[h]);
    }
    for (i=1;i<=n;i++)
    printf("%.0lf %.0lf
",f1[i],f2[i]);
    for (i=1;i<=n;i++)
    printf("%d
",(int)ceil(max(f1[i],f2[i]))-a[n-i+1]);
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[500005],n,b[500005];
double sq[500005],f1[500005],f2[500005];
void solve1(int l,int r,int L,int R)
{int i,Mid=0;
    if (l>r) return;
    int mid=(l+r)/2;
    double s=0;
    f1[mid]=a[mid];Mid=mid;
    for (i=L;i<=min(R,mid);i++)
    {
        s=a[i]+sq[mid-i];
        if (s>f1[mid]) f1[mid]=s,Mid=i; 
    }
    solve1(l,mid-1,L,Mid);
    solve1(mid+1,r,Mid,R);
}
void solve2(int l,int r,int L,int R)
{int i,Mid=0;
    if (l>r) return;
    int mid=(l+r)/2;
    double s=0;
    f2[n-mid+1]=a[mid];Mid=mid;
    for (i=L;i<=min(R,mid);i++)
    {
        s=a[i]+sq[mid-i];
        if (s>f2[n-mid+1]) f2[n-mid+1]=s,Mid=i; 
    }
    solve2(l,mid-1,L,Mid);
    solve2(mid+1,r,Mid,R);
}
int main()
{int i,j;
    scanf("%d",&n); 
    for (i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        sq[i]=sqrt((double)i);
    } 
    solve1(1,n,1,n);
    for (i=1;i<=n;i++)
    b[i]=a[n-i+1];
    for (i=1;i<=n;i++)
    a[i]=b[i];
    solve2(1,n,1,n);
    for (i=1;i<=n;i++)
    printf("%d
",(int)ceil(max(f1[i],f2[i]))-a[n-i+1]);
}