qwq斜率優化好題

第一步還是考慮最樸素的 $dp$

$dp=d$

p [ j ] + ( i − j − 1 + s u m [ i ] − s u m [ j ] ) 2 dp=dp[j]+(i-j-1+sum[i]-sum[j])^2

設 $f[i]=sum[i]+i$

那麼考慮將上述柿子變成 $dp[i]=dp[j]+(f[i]-f[j]-1-l)^2$

$= dp[j]+f[j]^2-2\times f[j]\times (2[i]-1) - 2\times l \times f[j]$

當存在一個 $j>k且j比k優秀的條件是$
$dp[j]+(f[i]-f[j]-1-l)^2 < dp[k]+(f[i]-f[k]-1-l)^2$

經過一波化簡
$\frac{dp[j]+f[j]^2-dp[k]-f[k]^2}{f[j]-f[k]} < 2\times (f[i]-l)$

然後直接套上斜率優化即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define mk make_pair
#define ll long long
#define int long long
using namespace std;
inline int read()
{
  int x=0,f=1;char ch=getchar();
  while (!isdigit(ch)) {if (ch=='-') f=-1;ch=getchar();}
  while (isdigit(ch)) {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
  return x*f;
}
const int maxn = 4e5+1e2;
struct Point
{
	int x,y,num;
}; 
Point q[maxn];
int n,m;
int sum[maxn];
int val[maxn];
int f[maxn];
int head=1,tail=0;
int dp[maxn];
int chacheng(Point x,Point y)
{
	return x.x*y.y-x.y*y.x;
}
bool count(Point i,Point j,Point k)
{
	Point x,y;
	x.x=k.x-i.x;
	x.y=k.y-i.y;
	y.x=k.x-j.x;
	y.y=k.y-j.y;
	if (chacheng(x,y)<=0) return true;
	return false;
}
void push(Point x)
{
	while (tail>=head+1 && count(q[tail-1],q[tail],x)) tail--;
	q[++tail]=x;
}
void pop(int lim)
{
   while (tail>=head+1 && q[head+1].y-q[head].y<lim*(q[head+1].x-q[head].x)) head++; 
} 
signed main()
{
  n=read();
  int l=read();
  for (int i=1;i<=n;i++) val[i]=read();
  for (int i=1;i<=n;i++) sum[i]=sum[i-1]+val[i];
  for (int i=1;i<=n;i++) f[i]=i+sum[i];
  push((Point){0,0,0});
  for (int i=1;i<=n;i++)
  {
  	 pop(2*(f[i]-l));
  	 int now = q[head].num;
  	 dp[i]=dp[now]+(f[i]-f[now]-1-l)*(f[i]-f[now]-1-l);
  	 push(Point{f[i],dp[i]+(f[i]+1)*(f[i]+1),i});
  	// cout<<i<<" "<<dp[i]<<endl;
  }
  cout<<dp[n]<<endl;
  return 0;
}