【BZOJ】【P1096】【ZJOI2007】【仓库建设】【题解】【斜率优化】

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=1096


我还是只会凸包+三分……

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;
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;
LL c[maxn],p[maxn],x[maxn],sump[maxn],sumxp[maxn];
int main(){
	n=getint();
	for(int i=1;i<=n;i++)x[i]=getint(),p[i]=getint(),c[i]=getint();
	for(int i=1;i<=n;i++)sumxp[i]=sumxp[i-1]+x[i]*p[i],sump[i]=sump[i-1]+p[i];//n++;
	C.push_back(point(0,0));
	f[1]=c[1];C.push_back(point(sump[1],f[1]+sumxp[1]));
	for(LL i=2;i<=n;i++){
		f[i]=1LL<<61;
		f[i]=C.Qmin(point(-x[i],1))+c[i]-sumxp[i-1]+x[i]*sump[i-1];
		C.push_back(point(sump[i],f[i]+sumxp[i]));	
	}cout<<f[n]<<endl;	
	return 0;
}


你可能感兴趣的:(bzoj)