Solution -「BalticOI 2004」Sequence

Description

http://222.180.160.110:1024/problem/28828.

Given is a sequencen \(A\) of \(n\) intergers.

Construct a stricly increasing sequence \(B\) of \(n\) intergers that makes the sum of \(|B_{i}-A_{i}|\) the smallest.

Solution

First, we make \(a_{i}:=a_{i}-i\). Then we just make "strictly increasing" become "unstrictly increasing".

1. for \(a_{1}\le a_{2}\le\cdots\le a_{n}\):

   When \(B\) is the same as \(A\), it gets the minimum answer.

2. for \(a_{1}\ge a_{2}\ge\cdots\ge a_{n}\):

   When for each \(i\), \(B_{i}=A_{\lfloor\frac{n}{2}\rfloor}\), it gets the minimum answer.

Maybe we can divide \(A\) into m parts.

Such like: \([l_{1},r_{1}],\cdots,[l_{m},r_{m}]\)

that satisfies: for each \(i\), sequence \(A[l_{i},r_{i}]\) is unstricly increasing/decreasing.

So we can get the answer to each interval.

Let's consider how to merge the answers.

When we're merging two intervals, we can just get the new median of the two intervals.

So things above are just bullshit.

Parallel Searching!

FUCK YOU.

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL INF=1e18;
int n;
LL a[1000010],b[1000010],ans;
void solve(LL l,LL r,int fr,int ba)
{
	if(l>=r||fr>ba)	return;
	LL mid=(l+r)>>1,tmp=0,mn=INF,pos=0;
	for(int i=fr;i<=ba;++i)	tmp+=abs(a[i]-mid-1);
	mn=tmp,pos=fr-1;
	for(int i=fr;i<=ba;++i)
	{
		tmp-=abs(a[i]-mid-1);
		tmp+=abs(a[i]-mid);
		if(tmp<mn)	mn=tmp,pos=i;
	}
	for(int i=fr;i<=pos;++i)	b[i]=mid;
	for(int i=pos+1;i<=ba;++i)	b[i]=mid+1;
	solve(l,mid,fr,pos),solve(mid+1,r,pos+1,ba);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)	scanf("%lld",&a[i]),a[i]-=i;
	solve(-INF,INF,1,n);
	for(int i=1;i<=n;++i)	ans+=abs(a[i]-b[i]);
	printf("%lld\n",ans);
	for(int i=1;i<=n;++i)	printf("%lld ",b[i]+i);
	return 0;
}

标签：le,just,Sequence,int,LL,Solution,ge,2004,increasing
来源： https://www.cnblogs.com/orchid-any/p/14490695.html