【题解】P3515 [POI2011]Lightning Conductor（二分栈/分治优化DP）

[POI2011]Lightning Conductor

题面翻译

给定一个长度为 $n$ 的序列 $\{a_n\}$，对于每个 $i\in [1,n]$ ，求出一个最小的非负整数 $p$ ，使得 $\forall j\in[1,n]$，都有 $a_j\le a_i+p-\sqrt{|i-j|}$

$1 \le n \le 5\times 10^{5}$，$0 \le a_i \le 10^{9}$ 。

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.

样例 #1

样例输入 #1

6
5
3
2
4
2
4

样例输出 #1

2
3
5
3
5
4

题解

先只考虑转移点小于当前点的情况。
画出转移点关于i的函数，可以发现两函数之间最多只有一个交点。
于此，可以得出二分栈做法或者决策单调性（四边形不等式也可以证）。
有决策单调性不要死证，可以画图辅助证明/打表/感性理解都是好的做法。

#include<bits/stdc++.h>
using namespace std;
inline int rd(){
	int f=1,j=0;
	char w=getchar();
	while(!isdigit(w)){
		if(w=='-')f=-1;
		w=getchar();
	}
	while(isdigit(w)){
		j=j*10+w-'0';
		w=getchar();
	}
	return f*j;
}
const int N=500010;
int n,sum[N];
double sq[N],f[N];
void deal(int l,int r,int L,int R){
	if(l>r)return ;
	int mid=(l+r)/2,p;
	double maxn=0;
	for(int i=L;i<=min(mid,R);i++){
		if(sq[mid-i]+sum[i]>maxn)maxn=sq[mid-i]+sum[i],p=i;
	}
//	cout<<mid<<":"<<p<<" "<<maxn<<"\n";
	f[mid]=max(f[mid],maxn);
	deal(l,mid-1,L,p),deal(mid+1,r,p,R);
	return ;
}
signed main(){
	n=rd();
	for(int i=1;i<=n;i++)sum[i]=rd(),sq[i]=sqrt(i),f[i]=0;
	deal(1,n,1,n);
	for(int i=1;i<=n/2;i++)swap(sum[i],sum[n-i+1]),swap(f[i],f[n-i+1]);
	deal(1,n,1,n);
	for(int i=n;i>=1;i--)printf("%d\n",(int)(ceil(f[i])-sum[i]));
	return 0;
}

双倍经验