数学专题训练6 计算几何

1. F-Girlfriend_2021牛客暑期多校训练营2 (nowcoder.com)

求两个球的交的体积

$|A{P}_1|\ge k_1|B{P}_1|,|C{P}_2|\ge k_2|D{P}_2|$​

简单积分就出来了.

#include
using namespace std;
const double PI = acos(-1);

struct P
{
    double x, y, z;

    double len()
    {
        return sqrt(x * x + y * y + z * z);
    }
};

P operator + (const P& a, const P& b)
{
    return {a.x + b.x, a.y + b.y, a.z + b.z};
}
P operator - (const P& a, const P& b)
{
    return {a.x - b.x, a.y - b.y, a.z - b.z};
}
P operator * (double a, const P& b)
{
    return {a * b.x, a * b.y, a * b.z};
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        P p1, p2, p3, p4, a, b, c, d;
        double k1, k2;

        scanf("%lf%lf%lf", &a.x, &a.y, &a.z);
        scanf("%lf%lf%lf", &b.x, &b.y, &b.z);
        scanf("%lf%lf%lf", &c.x, &c.y, &c.z);
        scanf("%lf%lf%lf", &d.x, &d.y, &d.z);
        scanf("%lf%lf", &k1, &k2);

        p1 = b + 1 / (k1 + 1) * (a - b);
        p2 = b + 1 / (k1 - 1) * (b - a);

        p3 = d + 1 / (k2 + 1) * (c - d);
        p4 = d + 1 / (k2 - 1) * (d - c);

        double r1 = (p1 - p2).len() / 2, r2 = (p3 - p4).len() / 2;
        P o1 = 0.5 * (p1 + p2), o2 = 0.5 * (p3 + p4);

        double dist = (o1 - o2).len();


        if(dist >= r1 + r2) printf("0\n");
        else if(dist <= fabs(r1 - r2))
        {
            double v1 = 4.0 / 3.0 * PI * r1 * r1 * r1;
            double v2 = 4.0 / 3.0 * PI * r2 * r2 * r2;
            printf("%.6f\n", min(v1, v2));
        }
        else
        {
            double cos1 = (r1 * r1 + dist * dist - r2 * r2) / (2 * r1 * dist);
            double cos2 = (r2 * r2 + dist * dist - r1 * r1) / (2 * r2 * dist);
            double sin1 = sqrt(1 - cos1 * cos1), sin2 = sqrt(1 - cos2 * cos2);
            double rr = r1 * sin1;
            double h1 = r1 * cos1, h2 = r2 * cos2;

            double v1 = 2.0 / 3.0 * PI * r1 * r1 * r1 * (1 - cos1) - 1.0 / 3 * PI * rr * rr * h1;
            double v2 = 2.0 / 3.0 * PI * r2 * r2 * r2 * (1 - cos2) - 1.0 / 3 * PI * rr * rr * h2;
            printf("%.6f\n", v1 + v2);
        }
    }
    return 0;
}