hdu4793Collision

2013年亚洲区域赛长沙站的几何题，就是告诉你两个圆（大圆和小圆），同心的，再告诉你一个小小圆在大圆外面，再告诉你小小圆的运动速度（矢量），如果碰到小圆就原路返回，如果碰不到就穿过大圆，问你小小圆在大圆中的时间。一开始忘了判断速度方向，如果反向，就为0s，wa了n遍。。。。

具体做法，求圆心到小小圆运动直线的最短距离，如果大于R+r就为0，如果小于R+r但大于Rm+r就根据勾股定理求出在大圆中一半的距离，除以速度再乘2就是时间了，如果小于等于Rm+r也是根据勾股定理求出在大圆中的距离，除以速度再乘2。

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#define eps 1e-8#define zero(x) (((x)>0?(x):(-x))<eps)using namespace std;double Rm,R,r;struct point{    double x,y;    point (double x=0,double y=0):x(x),y(y) {}};typedef point Vector;Vector operator +(Vector A, Vector B){    return Vector(A.x+B.x , A.y+B.y);}struct line{    point a,b;};double xmult(point p1,point p2,point p0){    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}double Dot(Vector A,Vector B){    return A.x*B.x+A.y*B.y;}point intersection(point u1,point u2,point v1,point v2){    point ret=u1;    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));    ret.x+=(u2.x-u1.x)*t;    ret.y+=(u2.y-u1.y)*t;    return ret;}point ptoline(point p,point l1,point l2)//点到直线距离{    point t=p;    t.x+=l1.y-l2.y,t.y+=l2.x-l1.x;    return intersection(p,t,l1,l2);}double  distance(point p1,point p2){    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}double disptoline(point p,line l){    return fabs(xmult(p,l.a,l.b))/distance(l.a,l.b);}int main(){    point v,s;    point st;    line stt;    while(~scanf("%lf%lf%lf%lf%lf%lf%lf",&Rm,&R,&r,&s.x,&s.y,&v.x,&v.y))    {        if(Dot(s,v)>=0)        {            printf("%lf\n",0.0);            continue;        }        stt.a=s,stt.b=s+v;        point tmp = ptoline(st,stt.a,stt.b);        double dis = distance(tmp,st);        double H = R+r,h = Rm+r;        double speed = sqrt((v.x*v.x)+(v.y*v.y));        {            if(dis>=H)            {                printf("%lf\n",0.0);            }            else            {                if(dis>=h)                {                    double dist = sqrt((H*H)-(dis*dis));                    printf("%.3lf\n",dist/speed*2.0);                }                else                {                    double dist = sqrt((H*H)-(dis*dis))-sqrt((h*h)-(dis*dis));                    printf("%.3lf\n",dist/speed*2.0);                }            }        }    }    return 0;}