[JSOI2008]球形空间产生器sphere

Posted 2020-11-30 Agakiss

Description

[JSOI2008]球形空间产生器sphere

Solution

以二维的时候为例
球面上三个点分别为((a_1,a_2))，((b_1,b_2))，((c_1,c_2))
设球心的点为((x_1,x_2))，球的半径为(T)
列式得
(left{ egin{aligned} (a_1-x_1)^2+(a_2-x_2)^2&=(a_1)^2-2a_1x_1+(x_1)^2+(a_2)^2-2a_2x_2+(x_2)^2&=T\ (b_1-x_1)^2+(b_2-x_2)^2&=(b_1)^2-2b_1x_1+(x_1)^2+(b_2)^2-2b_2x_2+(x_2)^2&=T\ (c_1-x_1)^2+(c_2-x_2)^2&=(c_1)^2-2c_1x_1+(x_1)^2+(c_2)^2-2c_2x_2+(x_2)^2&=T end{aligned} ight.)
联立方程
((a_1-x_1)^2+(a_2-x_2)^2=(b_1-x_1)^2+(b_2-x_2)^2=(c_1-x_1)^2+(c_2-x_2)^2)
暴力展开
((a_1)^2-2a_1x_1+(x_1)^2+(a_2)^2-2a_2x_2+(x_2)^2=(b_1)^2-2b_1x_1+(x_1)^2+(b_2)^2-2b_2x_2+(x_2)^2=(c_1)^2-2c_1x_1+(x_1)^2+(c_2)^2-2c_2x_2+(x_2)^2)
整理得
((a_1)^2-2a_1x_1+(a_2)^2-2a_2x_2=(b_1)^2-2b_1x_1+(b_2)^2-2b_2x_2=(c_1)^2-2c_1x_1+(c_2)^2-2c_2x_2)
也可以写成这个样子
(left{ egin{aligned} (a_1)^2-2a_1x_1+(a_2)^2-2a_2x_2&=(b_1)^2-2b_1x_1+(b_2)^2-2b_2x_2\ (b_1)^2-2b_1x_1+(b_2)^2-2b_2x_2&=(c_1)^2-2c_1x_1+(c_2)^2-2c_2x_2 end{aligned} ight.)
移来移去
(left{ egin{aligned} -2a_1x_1-2a_2x_2+2b_1x_1+2b_2x_2&=-(a_1)^2-(a_2)^2+(b_1)^2+(b_2)^2\ -2b_1x_1-2b_2x_2+2c_1x_1+2c_2x_2&=-(b_1)^2-(b_2)^2+(c_1)^2+(c_2)^2 end{aligned} ight.)
移来移去
(left{ egin{aligned} (2b_1-2a_1)x_1+(2b_2-2a_2)x_2&=-(a_1)^2-(a_2)^2+(b_1)^2+(b_2)^2\ (2c_1-2b_1)x_1+(2c_2-2b_2)x_2&=-(b_1)^2-(b_2)^2+(c_1)^2+(c_2)^2 end{aligned} ight.)
然后我们发现它成了一个(n)元一次方程组，于是我们来高斯消元就行了

Code