[ZJOI2007]仓库建设

题意：将序列分成若干段，区间 l,r有代价，求最小代价。

显然有二维方程式

(large f_i=minlbrace f_j+sum_{k=j+1}^{i}p_i(x_k-x_i) brace +c_i)

(large f_i=minlbrace f_j+sum_{k=j+1}^i P_ix_k -sum_{k=j+1}^i P_ix_i brace+c_i)

拆开中间的式子。

设(T_i=sum_{j=1}^{i}x_jp_j,P_i=sum_{j=1}^{i}x_j)

(large f_i=minegin{equation} left{ egin{array}{} f_j+x_i(P_i-P_j)+T_j-T_i end{array} ight}+c_i end{equation})

(f_j+x_i(P_i-P_j)+T_j-T_ile f_k+x_i(P_i-P_k)+T_k-T_i)

(f_j+x_iP_j+T_jle f_k+x_iP_k+T_k)

(f_j+T_j-(f_k+T_k)le x_i(P_j-P_k))

(huge frac{f_j+T_j -(f_k+T_k)}{(P_j-P_k)}< x_i)

然后就愉快的维护一个下凸壳即可。

/*
@Date    : 2019-07-31 13:54:47
@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+7;
int _x[N],p[N],c[N];
int T[N],P[N];
int f[N];
inline double slope(int j,int k)
{
	return 1.0*(f[j]+T[j]-f[k]-T[k])/(P[j]-P[k]);
}
signed main(void)
{
	#ifndef ONLINE_JUDGE
//	File("");
	#endif
	int n=gi;
	for(int i=1;i<=n;++i)_x[i]=gi,p[i]=gi,c[i]=gi;
	for(int i=1;i<=n;++i)
		T[i]=T[i-1]+_x[i]*p[i],
		P[i]=P[i-1]+p[i];
	static int Q[N*2];
	int l=1,r=1;
	for(int i=1;i<=n;++i)
	{
		while(l<r&&slope(Q[l],Q[l+1])<_x[i])++l;
		int j=Q[l];
		f[i]=f[j]+_x[i]*(P[i]-P[j])+T[j]-T[i]+c[i];
		while(l<r&&slope(Q[r],Q[r-1])>slope(i,Q[r]))--r;
		Q[++r]=i;
	}
	cout<<f[n];
	return 0;
}