[斜率优化]BZOJ 1911—— [Apio2010]特别行动队

题目传送门

解题思路

转移方程不难推出。

f[i]=max(f[j]+a∗(s[i]−s[j])2+b∗(s[i]−s[j])+c)

转化后可得

f[i]=max(f[j]+a∗s[j]2−b∗s[j]−2∗a∗s[i]∗s[j])+a∗s[i]2+b∗s[i]+c

y=f[j]+a∗s[j]2−b∗s[j]

x=s[j]

k=2∗a∗s[i]

因为斜率递减，所以维护一个上凸壳。

#include<cstdio>
#include<algorithm>
#define maxn 1000005
#define LL long long
using namespace std;
struct jz{
LL x,y;
jz(LL a=0,LL b=0):x(a),y(b){}
jz operator-(const jz &b){return jz(x-b.x,y-b.y);}
}que[maxn];
LL a,b,c,n,s[maxn],f[maxn],now,til,ans;
char nc(){
static char buf[100000],*l=buf,*r=buf;
if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);
if (l==r) return EOF;return *l++;
}
inline LL _read(){
LL num=0,f=1;char ch=nc();
while(ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=nc();}
while(ch>='0'&&ch<='9') num=num*10+ch-48,ch=nc();
return num*f;
}
LL getb(jz x,LL k){return x.y-k*x.x;}
LL check(jz x,jz y){return x.x*y.y-x.y*y.x;}
int main(){
freopen("exam.in","r",stdin);
freopen("exam.out","w",stdout);
n=_read();a=_read();b=_read();c=_read();
for (int i=1;i<=n;i++) s[i]=s[i-1]+_read();
now=til=1;que[1]=jz(0,0);
for (int i=1;i<=n;i++){
while(now<til&&getb(que[now],2*a*s[i])<getb(que[now+1],2*a*s[i])) now++;
f[i]=getb(que[now],2*a*s[i])+a*s[i]*s[i]+b*s[i]+c;
jz x(s[i],f[i]+a*s[i]*s[i]-b*s[i]);
while(til>1&&check(que[til]-que[til-1],x-que[til-1])>=0) til--;
if (now>til) now=til;que[++til]=x;
ans=max(ans,f[i]);
}
printf("%lld\n",ans);
}