qwq斜率优化好题
第一步还是考虑最朴素的(dp)
[dp=dp[j]+(i-j-1+sum[i]-sum[j])^2
]
设(f[i]=sum[i]+i)
那么考虑将上述柿子变成$$dp[i]=dp[j]+(f[i]-f[j]-1-l)^2$$
[= dp[j]+f[j]^2-2 imes f[j] imes (2[i]-1) - 2 imes l imes f[j]
]
当存在一个(j>k且j比k优秀的条件是)
[dp[j]+(f[i]-f[j]-1-l)^2 < dp[k]+(f[i]-f[k]-1-l)^2
]
经过一波化简
[frac{dp[j]+f[j]^2-dp[k]-f[k]^2}{f[j]-f[k]} < 2 imes (f[i]-l)
]
然后直接套上斜率优化即可
#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;
}