BZOJ 1911([Apio2010]特别行动队-斜率优化第一题)

1911: [Apio2010]特别行动队

Time Limit: 4 Sec   Memory Limit: 64 MB

Submit: 1684  

Solved: 706

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Description

Input

Output

Sample Input

4
-1 10 -20
2 2 3 4

Sample Output

9

HINT

这题是斜率优化第一题：)

总体来说

我们试图把方程写成f[i]重j和k转移来哪个优

只与kjk与i有关

考虑到转移时点的横坐标是递减的（A*s[i]） 所以维护凸壳各种轻松~

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define MAXN (1000000+10)
long long sqr(long long x){return x*x;}
int n;
long long a[MAXN],A,B,C,s[MAXN],f[MAXN];
long long calc(long long i,long long j)
{
    long long tmp=s[j]-s[i-1];
    return A*sqr(tmp)+B*tmp+C;
}
struct node
{
    long long x,y;
    int no;
    node(){}
    node(int _no):no(_no),x(A*s[_no]),y(f[_no]+A*sqr(s[_no])-B*s[_no]){}
    friend long double kk(node a,node b){return (long double)(b.y-a.y)/(b.x-a.x);}
}q[MAXN];
//friend long long operator*(node a,node b){return a.x*b.y-a.y*b.x;}
struct V
{
    long long x,y;
    V(node a,node b):x(b.x-a.x),y(b.y-a.y){}
    friend long long operator*(V a,V b){return a.x*b.y-a.y*b.x;}
};
int main()
{
    freopen("bzoj1911.in","r",stdin);
    scanf("%d%lld%lld%lld",&n,&A,&B,&C);
    s[0]=0;
    For(i,n) scanf("%lld",&a[i]),s[i]=s[i-1]+a[i];

    scanf("%lld%lld%lld",&A,&B,&C);
    f[0]=0,f[1]=calc(1,1);
    int head=0,tail=1;
    q[0]=node(0);
    q[1]=node(1);
//  cout<<calc(1,3)<<' '<<calc(1,2)<<' '<<calc(3,3)<<endl;
    Fork(i,2,n)
    {
        long double k=2*s[i];
        while (tail-head>0&&kk(q[head],q[head+1])<=k) head++;
//      cout<<q[head].no<<' ';
        f[i]=f[q[head].no]+calc(q[head].no+1,i);
        while (tail-head>0&&V(q[tail-1],q[tail])*V(q[tail],node(i))<0) tail--;
        q[++tail]=node(i);
    }
    cout<<f[n]<<endl;
    return 0;
}