题目链接

传送门

题意

告诉你圆锥的底部圆的半径和圆锥的高，再给你一个点的坐标及其运动向量，问你这个点什么时候会与这个圆锥相撞。

思路

比赛场上二分一直没过但是有人二分过了，今天再写这题想再试下二分，最后发现了自己的问题所在（可能这个点在\(check\)的时候已经穿过圆锥但是我的代码会把它当成还没达到，所以就会使得\(lb=mid\)）。最后按照公式求解过了。
圆锥表面方程为
\[
\begin{aligned}
\begin{cases}
\frac{h-z}{h}=\frac{R}{r}\x^2+y^2= R
\end{cases}
\end{aligned}
\]
然后设相撞时间为\(t\)，则点的坐标变成
\[
x=x_0+v_xt\y=y_0+v_yt\z=z_0+v_zt
\]
然后联立化简得到
\[
(v_x^2+v_y^2-\frac{r^2v_z^2}{h^2})t^2+2(v_xx_0+v_yy_0-\frac{z_0v_zr^2}{h^2}+\frac{v_zr^2}{h})t+x_0^2+y_0^2-\frac{r^2}{h^2}(h-z_0)^2=0
\]
求得
\[
t_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\t_2=\frac{-b-\sqrt{b^2-4ac}}{2a}
\]
最后再\(check\)是否恰好相撞取最小值即可。

代码实现如下

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 998244353;
const int maxn = 5000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int t;
double r, h, vx, vy, vz, x, y, z;

bool check(double t) {
    double xx = x + vx * t;
    double yy = y + vy * t;
    double zz = z + vz * t;
    if(zz < 0) return false;
    if(zz > h) return false;
    double R = (h - zz) * r / h;
    return (sqrt(xx * xx + yy * yy) - R) <= eps;
}

int main() {
    scanf("%d", &t);
    int icase = 0;
    while(t--) {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &r, &h, &x, &y, &z, &vx, &vy, &vz);
        if(z < 0 && sqrt(x * x + y * y) <= r * r) {
            printf("Case %d: %.7f\n", ++icase, abs(z) / abs(vz));
            continue;
        }
        double a = vx * vx + vy * vy  - r * r * vz * vz / h / h;
        double b = 2 * (vx * x + vy * y - z * vz * r * r / h / h + h * vz * r * r / h / h);
        double c = x * x + y * y - r * r / h / h * (h - z) * (h - z);
        double ans1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
        double ans2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);
        double ans = INF;
        if(check(ans1)) ans = min(ans, ans1);
        if(check(ans2)) ans = min(ans, ans2);
        printf("Case %d: %.7f\n", ++icase, ans);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/Dillonh/p/11196418.html