E - The Tower（2018CCPC吉林赛区）（计算几何）

E - The Tower（2018CCPC吉林赛区）（计算几何）

Time limit：1000 ms
Memory limit：262144 kB
judge：
HDU 6559
vjudge

Problem Description

The Tower shows a tall tower perched on the top of a rocky mountain. Lightning strikes, setting the building alight, and two people leap from the windows, head first and arms outstretched. It is a scene of chaos and destruction.

There is a cone tower with base center at (0, 0, 0), base radius r and apex (0, 0, h). At time 0 , a point located at $(x_0,y_0,z_0)$ with velocity $(v_x,v_y,v_z)$. What time will they collide? Here is the cone tower.

Input

The first line contains testcase number $T(T\le 1000)$, For each testcase the first line contains spaceseparated real numbers $r$ and $h(1\le r,h\le 1000)$ — the base radius and the cone height correspondingly.
For each testcase the second line contains three real numbers $x_0,y_0,z_0(0\le |x_0|,|y_0|,z_0\le 1000)$. For each testcase the third line contains three real numbers $v_x,v_y,v_z(1\le v_2^x+v_2^y+v_2^z\le 3\times 10^6)$. It is guaranteed that at time 0 the point is outside the cone and they will always collide.

Output

For each testcase print Case i : and then print the answer in one line, with absolute or relative error not exceeding $10^{-6}$

Sample Input

2
1 2
1 1 1
-1.5 -1.5 -0.5
1 1
1 1 1
-1 -1 -1

Sample Output

Case 1: 0.3855293381
Case 2: 0.5857864376

题意

给你一个圆锥，其底面的圆心的坐标是 $(0,0,0)$，再给你一个圆锥外的点的坐标和初速度，且保证他们俩一定会相撞，问你相撞的时间。（点在 $y$ 轴上方）。

题解

列方程，联立后求解。

因为点在 $y$ 轴上方，所以联立方程时不必考虑圆锥的底面。

列出圆锥上表面的表达式：
$$\{\begin{array}{c}{x}^2+y^2=R^2\\ \frac{r-R}{r}=\frac{z}{h}\end{array}$$
列出点的轨迹方程：
$$\{\begin{array}{l}x=x_0+v_{x}t\\ y=y_0+v_{y}t\\ z=z_0+v_{z}t\end{array}$$
联立消元：
$$(x_0+v_{x}t{)}^2+(y_0+v_{x}t{)}^2=[r-\frac{r(z_0+v_{z}t)}{h}{]}^2$$
化简后为：
$$(v_x^2+v_y^2-\frac{r^2{v}_z^2}{h^2})t^2+(2x_0{v}_x+2y_0{v}_y+\frac{2r^2{v}_z}{h}-\frac{2r^2{z}_0{v}_z}{h^2})t+x_0^2+y_0^2-r^2+\frac{2r^2{z}_0}{h}-\frac{r^2{z}_0^2}{h^2}=0$$
再次化简后得：
$$a{t}^2+bt+c=0$$
其中
$$\{\begin{array}{l}a=v_x^2+v_y^2-\frac{r^2{v}_z^2}{h^2}\\ b=2x_0{v}_x+2y_0{v}_y+\frac{2r^2{v}_z}{h}-\frac{2r^2{z}_0{v}_z}{h^2}\\ c=x_0^2+y_0^2-r^2+\frac{2r^2{z}_0}{h}-\frac{r^2{z}_0^2}{h^2}\end{array}$$
利用求解公式求出较小的那个值就是答案：
$$ans=\frac{-b-\sqrt{b^2-4ac}}{2a}$$

代码

#include <bits/stdc++.h>
using namespace std;

int T;
double r, h, x_0, y_0, z_0, vx, vy, vz;

int main() {
    
    
	while (cin >> T) {
    
    
		for (int i = 1; i <= T; ++i) {
    
    
			scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &r, &h, &x_0, &y_0, &z_0, &vx, &vy, &vz);
			double a = vx * vx + vy * vy - vz * vz*r*r / h / h;
			double b = 2 * x_0 * vx + 2 * y_0 * vy + 2 * r*r*vz / h - 2 * z_0*vz*r*r / h / h;
			double c = x_0 * x_0 + y_0 * y_0 - r * r + 2 * r*r*z_0 / h - r * r*z_0*z_0 / h / h;
			double ans = (-b - sqrt(b*b - 4 * a*c)) / 2 / a;
			printf("Case %d: %.10f\n", i, ans);
		}
	}
	return 0;
}