[ACM_几何] The Deadly Olympic Returns!!! （空间相对运动之最短距离）

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28235#problem/B

题目大意： 有两个同时再空间中匀速运动的导弹，告诉一个时间以及各自的初始坐标和该时间时的坐标，求运动过程中的最短距离

解题思路：  求出相对初位置、相对速度，则答案就是原点到射线型轨迹的距离，注意是射线！！！

//******************************************************************************* #include<iostream> #include<algorithm> #include<cmath> #include<stdio.h> using namespace std; #define eps 1e-8 int dcmp(double x){ if(fabs(x)<eps)return 0; else return x<0 ? -1:1; } struct Point3{ double x,y,z; Point3(double x=0,double y=0,double z=0):x(x),y(y),z(z){} }; bool operator==(const Point3& a,const Point3& b){ return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0 && dcmp(a.z-b.z)==0 ; } typedef Point3 Vector3; Vector3 operator+(Vector3 A,Vector3 B){ return Vector3(A.x+B.x,A.y+B.y,A.z+B.z); } Vector3 operator-(Vector3 A,Vector3 B){ return Vector3(A.x-B.x,A.y-B.y,A.z-B.z); } Vector3 operator*(Vector3 A,double p){ return Vector3(A.x*p,A.y*p,A.z*p); } Vector3 operator/(Vector3 A,double p){ return Vector3(A.x/p,A.y/p,A.z/p); } double Dot(Vector3 A,Vector3 B){return A.x*B.x+A.y*B.y+A.z*B.z;} double Length(Vector3 A){return sqrt(Dot(A,A));} double Angle(Vector3 A,Vector3 B){return acos(Dot(A,B)/Length(A)/Length(B));} //叉积 Vector3 Cross(Vector3 A,Vector3 B){ return Vector3(A.y*B.z-A.z*B.y,A.z*B.x-A.x*B.z,A.x*B.y-A.y*B.x); } //点p到射线AB的距离 double DDSXJL(Point3 p,Point3 A,Point3 B){ if(A==B)return Length(p-A); Vector3 v1=B-A,v2=p-A; if(dcmp(Dot(v1,v2))<0)return Length(v2); else return Length(Cross(v1,v2))/Length(v1); } //****************************************************************************** int main(){ Point3 now[2],fut[2],delta; int T;cin>>T; for(int kase=1;kase<=T;kase++){ int time;cin>>time; cin>>now[0].x>>now[0].y>>now[0].z; cin>>fut[0].x>>fut[0].y>>fut[0].z; cin>>now[1].x>>now[1].y>>now[1].z; cin>>fut[1].x>>fut[1].y>>fut[1].z; Point3 B;//坐标原点 delta=now[1]-now[0];//相对初位置 Vector3 speed=((fut[1]-now[1])-(fut[0]-now[0]));//相对速度 printf("Case %d: %.4lf\n",kase,DDSXJL(B,delta,delta+speed));//答案就是原点到轨迹的距离 }return 0; } //******************************************************************************* View Code