点和直线

点和直线

https://www.cnblogs.com/clrs97/p/4403197.html

需要实现动态加入一个直线，删除一个直线，和查询在平面上到所有直线距离之和最小的点的坐标

也就是要求这个东西的最小值：

\[\sum\left(\frac{\left|a_ix+b_iy+c_i\right|}{\sqrt{d_i}}\right)^2 \]

这是一个二元函数

我们考虑对于将x视作是常数，观察对应的y的取值。

\[\begin{eqnarray*}ans&=&\sum\left(\frac{\left|a_ix+b_iy+c_i\right|}{\sqrt{d_i}}\right)^2\\&=&\sum\frac{\left(a_ix+b_iy+c_i\right)^2}{d_i}\\&=&\sum\frac{\left(b_iy+a_ix+c_i\right)^2}{d_i}\\&=&\sum\left(\frac{b_i^2y^2}{d_i}+\frac{2b_i\left(a_ix+c_i\right)y}{d_i}+\frac{\left(a_ix+c_i\right)^2}{d_i}\right)\\&=&\sum\frac{b_i^2y^2}{d_i}+\sum\frac{2b_i\left(a_ix+c_i\right)y}{d_i}+\sum\frac{\left(a_ix+c_i\right)^2}{d_i}\\&=&y^2\sum\frac{b_i^2}{d_i}+2y\sum\frac{b_i\left(a_ix+c_i\right)}{d_i}+\sum\frac{\left(a_ix+c_i\right)^2}{d_i}\\&=&y^2\sum\frac{b_i^2}{d_i}+2y\left(\sum\frac{a_ib_ix}{d_i}+\sum\frac{b_ic_i}{d_i}\right)+\sum\frac{\left(a_ix+c_i\right)^2}{d_i}\\&=&y^2\sum\frac{b_i^2}{d_i}+2y\left(x\sum\frac{a_ib_i}{d_i}+\sum\frac{b_ic_i}{d_i}\right)+\sum\frac{\left(a_ix+c_i\right)^2}{d_i}\end{eqnarray*} \]

发现对应y的取值是一个二次函数，那么我们可以求出它的对称轴，就是取最小值的点。

\[\begin{eqnarray*}y&=&-\frac{2\left(x\sum\frac{a_ib_i}{d_i}+\sum\frac{b_ic_i}{d_i}\right)}{2\sum\frac{b_i^2}{d_i}}\\&=&-\frac{x\sum\frac{a_ib_i}{d_i}+\sum\frac{b_ic_i}{d_i}}{\sum\frac{b_i^2}{d_i}}\\&=&-\frac{\sum\frac{a_ib_i}{d_i}}{\sum\frac{b_i^2}{d_i}}x-\frac{\sum\frac{b_ic_i}{d_i}}{\sum\frac{b_i^2}{d_i}}\\&=&Ax+B\\A&=&-\frac{\sum\frac{a_ib_i}{d_i}}{\sum\frac{b_i^2}{d_i}}\\B&=&-\frac{\sum\frac{b_ic_i}{d_i}}{\sum\frac{b_i^2}{d_i}}\end{eqnarray*} \]

那么我们惊喜地发现y的取值是关于x的一个函数，所以把它直接带回去，就可以得到最优解

\[\begin{eqnarray*}ans&=&\sum\frac{\left(a_ix+b_iy+c_i\right)^2}{d_i}\\&=&\sum\frac{\left(a_ix+b_i\left(Ax+B\right)+c_i\right)^2}{d_i}\\&=&\sum\frac{\left(a_ix+Ab_ix+Bb_i+c_i\right)^2}{d_i}\\&=&\sum\frac{\left(\left(a_i+Ab_i\right)x+Bb_i+c_i\right)^2}{d_i}\\&=&\sum\frac{\left(a_i+Ab_i\right)^2x^2+2\left(a_i+Ab_i\right)\left(Bb_i+c_i\right)x+\left(Bb_i+c_i\right)^2}{d_i}\\&=&\sum\left( \frac{\left(a_i+Ab_i\right)^2x^2}{d_i}+\frac{2\left(a_i+Ab_i\right)\left(Bb_i+c_i\right)x}{d_i}+\frac{\left(Bb_i+c_i\right)^2}{d_i}\right)\\&=&x^2\sum\frac{\left(a_i+Ab_i\right)^2}{d_i}+2x\sum\frac{\left(a_i+Ab_i\right)\left(Bb_i+c_i\right)}{d_i}+\sum\frac{\left(Bb_i+c_i\right)^2}{d_i}\\&=&Ux^2+Vx+W\\U&=&\sum\frac{\left(a_i+Ab_i\right)^2}{d_i}\\&=&\sum\frac{a_i^2+2Aa_ib_i+A^2b_i^2}{d_i}\\&=&\sum\frac{a_i^2}{d_i}+2A\sum\frac{a_ib_i}{d_i}+A^2\sum\frac{b_i^2}{d_i}\\V&=&2\sum\frac{\left(a_i+Ab_i\right)\left(Bb_i+c_i\right)}{d_i}\\&=&2\sum\frac{Ba_ib_i+a_ic_i+ABb_i^2+Ab_ic_i}{d_i}\\&=&2\left(B\sum\frac{a_ib_i}{d_i}+\sum\frac{a_ic_i}{d_i}+AB\sum\frac{b_i^2}{d_i}+A\sum\frac{b_ic_i}{d_i}\right)\\W&=&\sum\frac{\left(Bb_i+c_i\right)^2}{d_i}\\&=&\sum\frac{B^2b_i^2+2Bb_ic_i+c_i^2}{d_i}\\&=&B^2\sum\frac{b_i^2}{d_i}+2B\sum\frac{b_ic_i}{d_i}+\sum\frac{c_i^2}{d_i}\end{eqnarray*} \]

然后我们发现最后的答案仍然是一个二次函数，所以我们只需要找出对称轴求解即可。

所以我们只需要对于每个直线维护6个变量，然后就可以方便加入删除了。