P3195 [HNOI2008]玩具装箱 题解

我数学好弱啊，这种题都能卡
设$f_i$为压缩完1~i的玩具的最优解，不难列出状态转移方程：
$f_i=\min \{f_j+[q_i-q_j+i-(j+1)-L{]}^2\}(j<i)$，其中$q_i={\sum }_{k=1}^i{C}_k$
明显是$O(n^2)$的式子，$n\le 5e4$这种数据肯定过不去。
就要用到一些奇怪的优化了，比如斜率优化。
设$s_i=q_i+i$，$L^{\prime }=L+1$
得$f_i=\min \{f_j+[(s_i-s_j)-L^{\prime }{]}^2\}(j<i)$
现在有两个数$j,k$，满足$i>j>k$，令状态$k$优于状态$j$，得
$$\begin{gathered} f_k+(s_i-s_k-L^{\prime }{)}^2\le f_j+(s_i-s_j-L^{\prime }{)}^2 \\ 遇见式子直接拆，得f_k+[s_i-(s_k+L^{\prime }){]}^2\le f_j+[s_i-(s_j+L^{\prime }){]}^2 \\ {f}_k+s_i^2-2s_i(s_k+L^{\prime })+(s_k+L^{\prime }{)}^2\le f_j+s_i^2-2s_i(s_j+L^{\prime })+(s_j+L^{\prime }{)}^2 \\ 化简，得f_k-2s_i(s_k+L^{\prime })+(s_k+L^{\prime }{)}^2\le f_j-2s_i(s_j+L^{\prime })+(s_j+L^{\prime }{)}^2 \\ 把有s_i项的移到一起，得-2s_i(s_k+L^{\prime })+2s_i(s_j+L^{\prime })\le -f_k+f_j-2s_i(s_j+L^{\prime })+2s_i(s_k+L^{\prime })-(s_k+L^{\prime }{)}^2+(s_j+L^{\prime }{)}^2 \\ 两边都乘-1，得2s_i(s_k+L^{\prime })-2s_i(s_j+L^{\prime })\le f_k-f_j+2s_i(s_k+L^{\prime })-2s_i(s_k+L^{\prime })+(s_k+L^{\prime }{)}^2-(s_j+L^{\prime }{)}^2 \\ {s}_i[2(s_k+L^{\prime })-2(s_j+L^{\prime })]\le f_k-f_j+2s_i(s_k+L^{\prime })-2s_i(s_k+L^{\prime })+(s_k+L^{\prime }{)}^2-(s_j+L^{\prime }{)}^2 \\ {s}_i\le \frac{f_k-f_j+2s_i(s_k+L^{\prime })-2s_i(s_k+L^{\prime })+(s_k+L^{\prime }{)}^2-(s_j+L^{\prime }{)}^2}{2(s_k+L^{\prime })-2(s_j+L^{\prime })} \\ 设X_i=2s_i,Y_i=f_i+(s_i+L^{\prime }{)}^2 \\ 得s_i\le \frac{Y_k-Y_j}{X_k-X_j} \\ 嗯，这么眼熟，高中书上的斜率？ \end{gathered}$$
图片来源：原题luogu题解

可以发现$s_i$是单调上升的，然后非常simple画个图像（直接到洛谷上去找吧），发现是下凸包，我们只要找第一个$slope(P_j,P_{j+1})>s_i$的点，那么既可以用二分$O(nlogn)$，也可以用单调队列$O(n)$，不过需要四边形不等式证明一下（口胡？不会），GG。
好像就结束了啊，放个代码：

# include 
using namespace std;
const int N=5e4+5;
int n,L;
long long q[N];
long long dp[N];
int Q[N<<2];
int front,rear;
long long sq(long long x) {return x*x;}
long long X(int i) 
{
	return 2.0*q[i];
}
long long Y(int i) 
{
	return (dp[i])+sq((q[i])+(L+1));
}
long double slope(int i,int j) 
{
	return (long double)(Y(j)-Y(i))/(X(j)-X(i));
}
int main(void) 
{
	scanf("%d%d",&n,&L);
	long long x;
	for(int i=1;i<=n;i++) 
	{
		scanf("%lld",&x);
		q[i]=q[i-1]+x+1;
	}
	front=rear=0;
	Q[++rear]=0;
	for(int i=1;i<=n;i++) 
	{
		while(front+1<rear&&slope(Q[front+1],Q[front+2])<=q[i]) front++;
		int head=Q[front+1];
		dp[i]=dp[head]+sq(q[i]-q[head]-L-1);
		while(front+1<rear&&slope(Q[rear-1],i)<slope(Q[rear],Q[rear-1])) rear--;
		Q[++rear]=i;
	}
	cout<<dp[n]<<endl;
	return 0;
}