BZOJ1010(HNOI2008)[玩具装箱toy]--斜率优化DP

【链接】
bzoj1010

【解题报告】

典型的斜率优化DP，但要证决策单调性

hzwer大佬写的证明(点这里)，博主好懒啊，QAQ

#include<cstdio>
#define LL long long
using namespace std;
const int maxn=50005;
int n,L,hed,til,que[maxn];
LL f[maxn],sum[maxn];
inline char nc()
{
    static char buf[100000],*l,*r;
    if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);
    if (l==r) return EOF; return *l++;
}
inline int Read()
{
    int res=0; char ch=nc();
    while (ch<'0'||ch>'9') ch=nc();
    while (ch>='0'&&ch<='9') res=res*10+ch-48,ch=nc();
    return res;
}
LL Sqr(LL x) {return x*x;}
LL Y(int k,int j) {return f[j]+Sqr(sum[j]+L)-f[k]-Sqr(sum[k]+L);}
LL X(int k,int j) {return 2*(sum[j]-sum[k]);}
int main()
{
    freopen("1010.in","r",stdin);
    freopen("1010.out","w",stdout);
    n=Read(); L=Read()+1; sum[0]=0;
    for (int i=1; i<=n; i++) sum[i]=sum[i-1]+Read()+1;
    hed=0; til=0; que[0]=0; f[0]=0;
    for (int i=1; i<=n; i++)
    {
        while (hed<til&&Y(que[hed],que[hed+1])<=X(que[hed],que[hed+1])*sum[i]) hed++;
        int j=que[hed]; f[i]=f[j]+Sqr(sum[i]-sum[j]-L);
        while (hed<til&&Y(que[til-1],que[til])*X(que[til],i)>=Y(que[til],i)*X(que[til-1],que[til])) til--;
        que[++til]=i;
    }
    printf("%lld\n",f[n]);
    return 0;
}