【BZOJ1010】玩具装箱

题解：

h[ ]单调，s[ ] 单调，所以直接单调队列维护一个上凸壳即可

//by sdfzchy
#include
#include
#include
#include
using namespace std;
typedef long long LL;
const int inf=(1<<30),N=1000010;
const double eps=1e-12;

inline int in()
{
    char ch=getchar();
    int f=1,tmp=0;
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') {tmp=(tmp<<1)+(tmp<<3)+(ch-'0');ch=getchar();}
    return tmp*f;
}

int n,m;
LL f[N],h[N],s[N],L;
inline LL gi(int x) {return f[x]+s[x]*s[x];}

inline double cal(int x,int y)
{
    return ((double)gi(y)-gi(x))/((double)s[y]-s[x]);
}
int q[N];

int main()
{
    n=in(),L=in();
    memset(f,0x3f,sizeof(f));
    for(int i=1;i<=n;i++)   s[i]=in()+s[i-1];
    for(int i=1;i<=n;i++)   s[i]+=i;
    for(int i=1;i<=n;i++)   h[i]=s[i]-(L+1);
    int l=1,r=0;  f[0]=0,q[++r]=0;
    for(int i=1;i<=n;i++)
    {
        while(lq[l],q[l+1])<2*h[i]) l++;
        int k=q[l]; f[i]=f[k]+(s[i]-s[k]-(L+1))*(s[i]-s[k]-(L+1));
        while(lq[r-1],q[r])-cal(q[r],i)>eps) r--;
        q[++r]=i;
    }
    printf("%lld\n",f[n]);
    return 0;
}