about_sqrt_and_acos
关于sqrt对负数开根号acos对大于1的数计算反三角函数值的问题
空间里面有一个实心球,两个点,问两点不经过实心球的路径的最小值.
题目意思很简单,我们可能会用到余弦定理,但是,余弦定理有误差,我们可能会得到一些奇怪的数字,浮点数的误差导致
余弦定理计算出来了小于-1或者大于1的其他数字,当我们对这样的数字进行反三角函数运算时,会得到nan,当然sqrt
有时候也会碰到对负数开方,于是我们要重写这两个函数
#include<bits/stdc++.h> using namespace std; const double eps = 1e-12; const double INF = 1e18; const double PI = acos(-1.0); double ACOS(double x) { if(x>1) x=1.0; if(x<-1) x=-1.0; return acos(x); } double SQRT(double x) { if(x<0) return 0; return sqrt(x); } struct Point3D { double x, y, z; Point3D() {}; Point3D(double xx, double yy,double zz) { x = xx; y = yy; z = zz; } }; typedef struct Point2D { Point2D() {}; double x; double y; Point2D(double xx, double yy) { x = xx; y = yy; } } Point; struct Circle3D { Point3D o; double r; }; struct Circle { Point2D o; double r; }; struct Line { Point s, e; Line() {}; Line(Point a, Point b) { s = a; e = b; }; }; double dis(Point a, Point b) { return SQRT(pow(a.x - b.x, 2) + pow(a.y - b.y, 2)); } double dis(Point3D a, Point3D b) { return SQRT(pow(a.x - b.x, 2) + pow(a.y - b.y, 2) + pow(a.z - b.z, 2)); } double solve(Point3D s, Point3D t, Circle3D o) { double so=dis(s,o.o); double to=dis(t,o.o); double st=dis(s,t); double sita_sot=ACOS((so*so+to*to-st*st)/2/so/to); double sita_sto=ACOS((st*st+to*to-so*so)/2/st/to); double sita_tso=ACOS((st*st+so*so-to*to)/2/st/so); double p=(so+to+st)/2; double d=SQRT(p*(p-so)*(p-to)*(p-st))*2/st; if(sita_sto<PI/2&&sita_tso<PI/2&&d<o.r) { double sita_soa1=ACOS(o.r/so); double sita_toa2=ACOS(o.r/to); double sita=sita_sot-sita_soa1-sita_toa2; return sita*o.r + so*sin(sita_soa1) + to*sin(sita_toa2); } else { return dis(s, t); } } double solve(Point s, Point t, Circle o) { double so=dis(s,o.o); double to=dis(t,o.o); double st=dis(s,t); double sita_sot=ACOS((so*so+to*to-st*st)/2/so/to); double sita_sto=ACOS((st*st+to*to-so*so)/2/st/to); double sita_tso=ACOS((st*st+so*so-to*to)/2/st/so); double p=(so+to+st)/2; double d=SQRT(p*(p-so)*(p-to)*(p-st))*2/st; if(sita_sto<PI/2&&sita_tso<PI/2&&d<o.r) { double sita_soa1=ACOS(o.r/so); double sita_toa2=ACOS(o.r/to); double sita=sita_sot-sita_soa1-sita_toa2; return sita*o.r + so*sin(sita_soa1) + to*sin(sita_toa2); } else { return dis(s, t); } } double solve(double a,double b,double c,double r) { Point s, t; Circle o; double arcsita = (pow(b, 2) + pow(c, 2) - pow(a, 2)) / 2 / b / c; double sita = ACOS(arcsita); t.x = c * cos(sita); t.y = c * sin(sita); o.o.x = 0; o.o.y = 0; o.r = r; s.x = b; s.y = 0; return solve(s, t, o); } int main() { int T; scanf("%d", &T); while (T--) { Circle3D o; Point3D s, t; scanf("%lf%lf%lf%lf", &o.o.x, &o.o.y, &o.o.z,&o.r); scanf("%lf%lf%lf", &s.x, &s.y, &s.z); scanf("%lf%lf%lf", &t.x, &t.y, &t.z); double a = dis(s, t); double b = dis(s, o.o); double c = dis(t, o.o); if(fabs(a)<eps) { printf("0.0000000000\n"); } else { printf("%.8lf\n", solve(a,b,c,o.r)); } } }
本博客所有文章除特别声明外,均采用 CC BY-NC-SA 4.0 许可协议。转载请注明来自 Believe it!