传送门
第一次自己推斜率优化,好高兴
对于区间的长度,s[i]表示前缀和在加上i
也就是
![s[i]=i+\sum a[j](1<=j<=i) 玩具装箱TOY[斜率优化]_#define](https://file.cfanz.cn/uploads/gif/2022/07/12/8/5CG0ccLbY5.gif)
首先,考虑暴力的DP
![f[i] = min ( f[j] + (s[i]-s[j]-L-1)^2 ) 玩具装箱TOY[斜率优化]_i++_02](https://file.cfanz.cn/uploads/gif/2022/07/12/8/2225E5E601.gif)
拆开
![f[i] = min ( f[j] + s[i]^2 + s[j]^2 + L^2 -2*s[i]*s[j] - 2*s[i]*L + 2*s[j]*L -2s[i] + 2s[j] + 2L +1) 玩具装箱TOY[斜率优化]_前缀和_03](https://file.cfanz.cn/uploads/gif/2022/07/12/8/d960DVU9b3.gif)
整理成如下形式
也就是
![f[i] = min ( f[j] + s[j]^2 + 2*s[j] - 2*(s[i]-L)*s[j] + ...) 玩具装箱TOY[斜率优化]_i++_05](https://file.cfanz.cn/uploads/gif/2022/07/12/8/S1WPcL1B55.gif)
最小值维护下凸包
#include<bits/stdc++.h>
#define y(A) (f[A]+s[A]*s[A]+s[A])
#define k(A) (2*s[A]-2*L)
#define x(A) (s[A])
#define N 50005
#define inf 1000000000000000
#define LL long long
using namespace std;
LL n,L,s[N],f[N],q[N];
double K(int i,int j){return 1.0*(y(j)-y(i))/(x(j)-x(i));}
int main(){
scanf("%lld%lld",&n,&L);
for(int i=1;i<=n;i++){
LL x; scanf("%lld",&x);
s[i]=s[i-1]+x;
}
for(int i=1;i<=n;i++) s[i]+=i;
int l=1,r=1;
for(int i=1;i<=n;i++){
while(l<r && K(q[l],q[l+1]) <= k(i)) l++;
f[i] = f[q[l]] + (s[i]-s[q[l]]-L-1)*(s[i]-s[q[l]]-L-1);
while(l<r && K(q[r],i) <= K(q[r-1],q[r])) r--;
q[++r] = i;
}printf("%lld",f[n]); return 0;
}










