2013 ACM区域赛长沙 C Collision HDU 4793

题意：在平面上0,0点，有一个半径为R的圆形区域，并且在0,0点固定着一个半径为RM（<R）的圆形障碍物，现在圆形区域外x,y，有一个半径 为r的，并且速度为vx,vy的硬币，如果硬币碰到了障碍物，将会保持原有的速度向反射的方向继续前进，现在给出R,RM,r,x,y,vx,vy，问硬币的任意部分在圆形区域中滑行多少时间？

思路：首先把R,RM加上r，就可以把硬币看做一个点来讨论了，然后算一下射线与这两个圆交点的个数，记作C1，C2，CM1，CM2,若与两个圆的交点数都是2，答案就是dis(c1,c2)-dis(cm1,cm2)之后再除以合成后的速度，若与大圆的交点数为2，小圆的交点数是0或者1，答案就是 dis（c1,c2）/速度，否则肯定是0了.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<cmath>
#include<vector>
#define inf 0x3f3f3f3f
#define Inf 0x3FFFFFFFFFFFFFFFLL
#define eps 1e-9
#define pi acos(-1.0)
using namespace std;
typedef long long ll;

int dcmp(double x) {
    return (x>eps)-(x<-eps);
}

typedef struct Point {
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}
    Point operator+(const Point& p) const {
        return Point(x+p.x, y+p.y );
    }
    Point operator-(const Point& p) const {
        return Point(x-p.x, y-p.y );
    }
    Point operator*(const double d) const {
        return Point(x*d, y*d );
    }
    Point operator/(const double d) const {
        return Point(x/d, y/d );
    }
    void read() {
        scanf("%lf%lf", &x, &y);
    }
} Vector;

inline double Dis(const Point& a, const Point& b) {
    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}

struct Line {
    Point P;
    Vector v;
    double ang;
    Line() {}
    Line(const Point& P, const Vector& v):P(P),v(v) {
        ang = atan2(v.y,v.x);
    }
    bool operator<(const Line& L) const {
        return ang<L.ang;
    }
    Point point(double t) {
        return P+v*t;
    }
};

struct Circle {
    Point c;
    double r;
    Circle() {}
};

int GetLineCircleIntersection(Line L, Circle C, double &t1, double &t2, vector<Point> &sol) {
    double a = L.v.x, b = L.P.x-C.c.x, c = L.v.y, d = L.P.y - C.c.y;
    double e = a*a+c*c, f = 2*(a*b+c*d), g = b*b+d*d-C.r*C.r;
    double delta = f*f-4*e*g;
    if(dcmp(delta)<0) return 0;
    if(dcmp(delta)==0) {
        t1 = t2 = -f/(2*e);
        sol.push_back(L.point(t1));
        return 1;
    }
    t1 = (-f-sqrt(delta))/(2*e);
    sol.push_back(L.point(t1));
    t2 = (-f+sqrt(delta))/(2*e);
    sol.push_back(L.point(t2));
    if(dcmp(t1)<0 || dcmp(t2)<0) return 0;
    return 2;
}

double R,RM,r,x,y,vx,vy;
int n,m,k;

int main() {
//    freopen("in.txt","r",stdin);
    while(~scanf("%lf%lf%lf%lf%lf%lf%lf",&RM,&R,&r,&x,&y,&vx,&vy)) {
        Line l1(Point(x,y),Point(vx,vy));
        Circle c1,c2;
        c1.r=RM+r;
        c1.c.x=c1.c.y=0.0;
        c2.c.x=c2.c.y=0.0;
        c2.r=R+r;
        double db1=0.0,db2=0.0;
        vector<Point> crs1;
        vector<Point> crs2;
        int num2=GetLineCircleIntersection(l1,c1,db1,db2,crs2);
        int num1=GetLineCircleIntersection(l1,c2,db1,db2,crs1);
//        printf("num1:%d num2:%d
",num1,num2);
        double len=0.0,ans=0.0;
        double vv=sqrt(vx*vx+vy*vy);
        if (num1==2 && num2==2) {
            len=Dis(crs1[0],crs1[1])-Dis(crs2[0],crs2[1]);
            printf("%.4lf
",len/vv);
        } else if (num1==2) {
            len=Dis(crs1[0],crs1[1]);
            printf("%.4lf
",len/vv);
        } else {
            printf("0.0000
");
        }
    }
    return 0;
}