【HNOI2008】玩具装箱

\(ProblemLink\)

设\(S_i=\sum_{i=1}^n C_i\)

易得\(n^2\)方程

设\(f[i]\)为处理前\(i\)个玩具的最小费用

\(f_i=min(f_j+(i-j+s_i-s_j-L)^2)(j<i)\)

\(f_i=f_j+(i-j+s_i-s_j-L-1)^2\)

设\(a_i=s_i+i,b_j=-s_j-j-L-1\)

\(f_i=f_j+a_i^2+2a_ib_j+b_j^2\)

将\(i\),\(j\)相关的放到一起

\(f_i-a_i^2-2a_ib_j=f_j+b_j^2\)

设\(b_j=x,f_j+b_j^2=y,f_i-a_i^2=b\)

则变为一条直线\(-2a_ix+b=y\)

我们实际上要找的答案\(f_i\)就是最小化一个斜率为\(-2a_i\)的直线过点\((x,y)\)在\(y\)轴上的截距\(b\)加上定值\(a_i^2\)

因为\(-2a_i\)递减,用单调队列维护一个上凸包就可以了。

/*
@Date    : 2019-07-31 10:46:24
@Author  : Adscn (adscn@qq.com)
@Link    : https://www.cnblogs.com/LLCSBlog
*/
#include<bits/stdc++.h>
using namespace std;
#define IL inline
#define RG register
#define gi getint()
#define gc getchar()
#define File(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
IL int getint()
{
	RG int xi=0;
	RG char ch=gc;
	bool f=0;
	while(ch<'0'||ch>'9')ch=='-'?f=1:f,ch=gc;
	while(ch>='0'&&ch<='9')xi=(xi<<1)+(xi<<3)+ch-48,ch=gc;
	return f?-xi:xi;
}
template<typename T>
IL void pi(T k,char ch=0)
{
	if(k<0)k=-k,putchar('-');
	if(k>=10)pi(k/10,0);
	putchar(k%10+'0');
	if(ch)putchar(ch);
}
#define int long long
const int N=1e6;
int s[N];
int f[N];
int n,L;
inline int A(int i){return s[i]+i;}
inline int B(int i){return -s[i]-i-L-1;}
inline int sqr(int i){return i*i;}
inline double slope(int i,int j){return (double)(f[i]+sqr(B(i))-f[j]-sqr(B(j)))/(B(i)-B(j));}
signed main(void)
{
	n=gi,L=gi;
	for(int i=1;i<=n;++i)s[i]=s[i-1]+gi;
	static int Q[N*2];
	int l=0,r=-1;
	Q[++r]=0;
	for(int i=1;i<=n;++i)
	{
		while(l<r&&slope(Q[l],Q[l+1])>=-2*A(i))++l;
		f[i]=f[Q[l]]+sqr(A(i)+B(Q[l]));
	//	assert(slope(1,2)==slope(2,1));
		while(l<r&&slope(Q[r-1],Q[r])<=slope(Q[r-1],i))--r;
		Q[++r]=i;
	}
	cout<<f[n]<<endl;
	return 0;
}