【BZOJ】【P3156】【防御准备】【题解】【斜率优化】

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=3156

dp方程:

reverse(a)

f[1]=a[1]

f[i]=a[i]+min(f[j]+(i-j)*(i-j-1)/2)

f[i]=a[i]+(i*i-i)/2+min(f[j]+(j*j+j)/2-ij)

是斜率优化

呃,我只会nlogn的凸包+三分怎么破……

Code:

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e6+5;
struct point{
	LL x,y;
	point(LL _x=0,LL _y=0):x(_x),y(_y){}
	LL operator*(point oth)const{return x*oth.y-y*oth.x;}
	LL operator^(point oth)const{return x*oth.x+y*oth.y;}
	point operator-(point oth)const{return point(x-oth.x,y-oth.y);}
};
LL n,a[maxn];
LL f[maxn];
LL getint(){
	LL res=0;char c=getchar();
	while(!isdigit(c))c=getchar();
	while(isdigit(c))res=res*10+c-'0',c=getchar();
	return res;
}
struct CH{
	point ch[maxn];
	int m;
	CH():m(0){}
	void push_back(point p){
		while(m>1&&(ch[m-1]-ch[m-2])*(p-ch[m-1])<=0)m--;
		ch[m++]=p;
	}
	LL Qmin(point p){
		int l=0,r=m-1;
		while(r-l>2){
			int mid1=l+(r-l)/3;
			int mid2=r-(r-l)/3;
			if((p^ch[mid1])<(p^ch[mid2]))
				r=mid2;
			else l=mid1;
		}LL ans=1LL<<61;
		for(int i=l;i<=r;i++)ans=min(ans,p^ch[i]);
		return ans;
	}
}C;
int main(){
	n=getint();
	for(LL i=1;i<=n;i++)a[i]=getint();
	reverse(a+1,a+1+n);n++;
	f[1]=a[1];C.push_back(point(1,f[1]+1));
	for(LL i=2;i<=n;i++){
		f[i]=1LL<<61;
		f[i]=C.Qmin(point(-i,1))+(LL)a[i]+(LL)((LL)i*i-i)/2;
		C.push_back(point(i,f[i]+(LL)(i*i+i)/2));	
	}cout<<min(f[n],f[n-1])<<endl;	
	return 0;
}