Optimization slope \ (the DP \) Luanjiang
About slope optimization, I'm a fool ah, I have been really did not understand ......
Write something to help understand it ...... total of two examples, are quite typical, I just go into detail first, the other would talk a little about it.
\ (T1 [HNOI2018] TOY \ ) Toys Packing
And the equation of state is very good out of ah:
\ (f [i] = min (f [j] + (s [i] -s [j] + ijL-1) ^ 2) \) where \ (s [i] \) represents the prefix and, \ (F [I] \) representing the forward \ (I \) the minimum value after treatments.
However, we found this thing if you want to transfer a \ (O (n ^ 2) \) and \ (N <= 50000 \) , apparently not transferred.
The next step is to optimize the slope of the world!
We are looking for is in the \ (i \) a preceding \ (j \) so that \ (f [i] \) minimum, consider how you can quickly find.
令\(a[i]=s[i]+i,b[i]=s[i]+i-L-1\)。
We deformation of the original equation, we can get this form:(Why Do not pipe Okay)
\ (2 * A [I] * B [J] + F [I] -a [I] = F ^ 2 [J] + B [J] ^ 2 \) . To be clear, when we transfer \ (i \) when \ (a [i] \) is determined.
Corresponding to about \ (KX + B = Y \) , we let \ (2 * a [i] \) of \ (K \) , \ (B [J] \) of \ (X \) , \ (F [ j] + b [j] ^ 2 \) of \ (Y \) , then we ask for the straight line \ (Y \) minimum on the intercept.
When we transfer \ (I \) , the foregoing \ (1 ~ i-1 \ ) may be represented as a stack point ( \ (P_j (B [J], F [J] + B [J] ^ 2) \) ).
It is the transfer, through a looking \ (P_j \) a slope of \ (2 * a [i] \) of a straight line, so that it \ (Y \) intercept the axis is minimized.
Wish to point out three, which constitutes an upwardly convex hull (e.g., upper right panel), we found that when we straight lines are translated, \ (B \) must not optimal decision.
So we can give it \ (B \) points, i.e., to maintain a downward convex hull (as shown in Figure) which we found in this case \ (A, B, C \ ) three points are likely to be the most excellent decision.
At this time, all considered maintenance may constitute a downwardly convex hull of the points with a data structure.
//回来看下之前我们说的斜率\(2*a[i]\)，显然它具有单调性（如果没有单调性就二分）
在前\(1～i-1\)中，不妨设有两个状态：\(j,k\)且\(j\)比\(k\)更优，则有：
\(f[j]+(s[i]-s[j]+i-j-L-1)^2<f[k]+(s[i]-s[k]+i-k-L-1)^2\)
经过一系列毫不人道的化简（风骨傲天很懒所以他没把将过程放上来）可以得到：
\(2*a[i]>\frac{f[j]-f[k]+b[j]^2-b[k]^2}{b[j]-b[k]}\)（好丑的式子……）
也就是说只要满足这个式子，就有\(j\)比\(k\)更优。
同样对应斜率，就有：\(P_j\)和\(P_k\)的斜率小于\(2*a[i]\)时，就有\(j\)比\(k\)更优。
又因为我们维护的是一个向下的凸包，所以我们就只用考虑相邻的两个点即可。
那么我们就可以用一个单调队列来维护，队列中相邻点间连线的斜率递增。
这张图应该说很好的反应了转移和维护的过程，上面的两张就是转移，下面的两张是维护。
转移：由图中我们可以发现，我们要求的\(j\)为斜率第一个大于\(2*a[i]\)的点，因此舍掉\(A,B\)
维护：因为单桥队列中斜率的单调性，删掉\(E\)，由转移后的\(i\)得出的点\(F(b[i],f[i]+b[i]^2)\)加入队尾。
而实际上，我们的斜率优化有一定的公式：
当方程形如：\(f[i]=min(f[j]+S(i,j))+k\)（\(k\)为常数）时，可以使用斜率优化。我们的斜率为在当次转移中的一个不变量，维护的是一堆点。不过通常用上面讲到的“假设两个状态法”来求斜率。
所以斜率优化的思考过程应该是和上面讲到的相反，先考虑斜率再变形方程并得出\(x,y\)。
啊～终于打完了，累成狗
又是快乐的代码时间：
#include<bits/stdc++.h>
using namespace std;
inline int read()
{
int f=1,w=0;char x=0;
while(x<'0'||x>'9') {if(x=='-') f=-1; x=getchar();}
while(x!=EOF&&x>='0'&&x<='9') {w=(w<<3)+(w<<1)+(x^48);x=getchar();}
return w*f;
}
const int N=50010;
int n,q[N],l,r,L;
double s[N],f[N];
inline double A(int i) {return s[i]+i;}
inline double B(int i) {return s[i]+i+L+1;}
inline double S(double x) {return x*x;}
inline double K(int i,int j) {return (f[i]-f[j]+S(B(i))-S(B(j)))/(B(i)-B(j));}
int main(){
#ifndef ONLINE_JUDGE
freopen("A.in","r",stdin);
#endif
n=read();L=read();l=r=1;
for(int i=1,c;i<=n;i++) c=read(),s[i]=c*1.0+s[i-1];
for(int i=1;i<=n;i++)
{
while(l<r&&2*(s[i]+i)>K(q[l],q[l+1])) l++;
f[i]=f[q[l]]+S(A(i)-B(q[l]));
while(l<r&&K(q[r-1],q[r])>K(i,q[r-1])) r--;
q[++r]=i;
}
printf("%lld",(long long)f[n]);
}
这一题也是一个经典的斜率优化，稍微有点不同。
状态：\(f[i]=min(f[j]+a*(s[i]-s[j])^2+b*(s[i]-s[j])+c)\)
化简：\(2*a*s[i]*s[j]+f[i]-a*s[i]^2-b*s[i]-c=f[j]+a*s[j]^2-b*s[j]\)
因为这一题的\(a<0\)，所以我们维护一个向上的凸包即可。
直接上代码：（我说过了，这个题解只会简单说明哦\(QwQ\)）
#include <cstdio>
using namespace std;
#define F(x) ((x)*(x))
inline int read()
{
int f=1,w=0;char x=0;
while(x<'0'||x>'9') {if(x=='-') f=-1; x=getchar();}
while(x!=EOF&&x>='0'&&x<='9') {w=(w<<3)+(w<<1)+(x^48);x=getchar();}
return w*f;
}
const int N=1000010;
#define int long long
int n,l,r,f[N],Q[N],a,b,c,s[N];
inline double Work(int x,int y)
{
return 1.*(f[x]-f[y]+(F(s[x])-F(s[y]))*a)/(s[x]-s[y])-b;
}
main(){
n=read(),a=read(),b=read(),c=read();
for(int i=1;i<=n;i++) s[i]=read(),s[i]+=s[i-1];
for(int i=1;i<=n;i++)
{
while(l<r&&Work(Q[l+1],Q[l])>s[i]*2*a) l++;
//用于从队列中选出最值
f[i]=f[Q[l]]+a*F(s[i]-s[Q[l]])+b*(s[i]-s[Q[l]])+c;
while(l<r&&Work(Q[r],Q[r-1])<Work(i,Q[r])) r--;
//维护队列单调性
Q[++r]=i;
}
printf("%lld",f[n]);
}