[HDU 5572] An Easy Physics Problem （点在线上判定+对称）

HDU - 5572

给定一个圆和圆外两个点 A和 B
现在有一个质点在 A处，有速度方向 V
其与圆的碰撞是弹性碰撞，问质点是否能经过 B

---

分情况讨论

1. 如果射线不与圆相交，直接判定点是否在射线上
2. 如果射线与圆相交，那么列方程解出与原交点
   并得出反弹的法线方程，然后以法线方程作对称
   最后判断点是否在一条线段和一条射线上

作对称的话可以将点 A以法线作对称，然后再用撞击点和对称点得到撞后轨迹

#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <iostream>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <cctype>#include <map>#include <set>#include <queue>#include <bitset>#include <string>#include <complex>using namespace std;typedef pair<int,int> Pii;typedef long long LL;typedef unsigned long long ULL;typedef double DBL;typedef long double LDBL;#define MST(a,b) memset(a,b,sizeof(a))#define CLR(a) MST(a,0)#define SQR(a) ((a)*(a))#define PCUT puts("\n----------")const DBL eps = 1e-8;int sgn(DBL x){return x<-eps?-1:(x>eps?1:0);}struct Vector{    DBL x, y;    Vector(DBL _x=0, DBL _y=0):x(_x),y(_y){}    Vector operator + (const Vector &rhs) const {return Vector(x+rhs.x, y+rhs.y);}    Vector operator - (const Vector &rhs) const {return Vector(x-rhs.x, y-rhs.y);}    Vector operator * (const DBL &rhs) const {return Vector(x*rhs, y*rhs);}    Vector operator / (const DBL &rhs) const {return Vector(x/rhs, y/rhs);}    DBL operator * (const Vector &rhs) const {return x*rhs.x + y*rhs.y;}    DBL operator ^ (const Vector &rhs) const {return x*rhs.y - y*rhs.x;}    int read(){return scanf("%lf%lf", &x, &y);}    int pri(){cout << x << " " << y << endl; return 0;}};typedef Vector Point;struct Line{    Vector u,v,s;    Line(){}    Line(Point _u, Point _v):u(_u),v(_v),s(_v-_u){}};struct Circle{    Point o; DBL r;    Circle(){};    void read(){o.read(); scanf("%lf", &r);}};int N;Point A, B, V;Circle O;DBL Vlen(Vector);Vector Normal(Vector);Point SISC(Line,Circle);bool PointOnHline(Point,Line);bool PointOnSegment(Point,Line);DBL PointToLine(Point,Line);int main(){    #ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);    #endif    int T;    scanf("%d", &T);    for(int ck=1; ck<=T; ck++)    {        O.read(); A.read(); V.read(); B.read();        Line l1(A, A+V);        DBL dist = PointToLine(O.o, l1);        bool ok=0;        if(sgn(dist-O.r)>=0 || sgn(V*(O.o-A))<=0) ok |= PointOnHline(B, l1);        else        {            Line l2(O.o, SISC(l1,O));            if(PointOnHline(O.o, l1)) ok |= PointOnHline(B, l2);            else            {                Vector n = Normal(l2.s);                if(sgn(n*V)<0) n=n*(-1);                n = n/Vlen(n);                ok |= PointOnSegment(B, Line(A, l2.v)) || PointOnHline(B, Line(l2.v, A+n*2*PointToLine(A, l2)));            }        }        printf("Case #%d: %s\n", ck, ok?"Yes":"No");    }    return 0;}DBL Vlen(Vector v) { return sqrt(SQR(v.x) + SQR(v.y));}Vector Normal(Vector v) {return Vector(-v.y, v.x);}bool PointOnHline(Point p, Line l){    Vector temp = p-l.u;    return sgn(temp^l.s) == 0 && sgn(temp*l.s) >= 0;}bool PointOnSegment(Point p, Line l){    return sgn( (l.u-p)^(l.v-p) )==0 && sgn( (l.u-p)*(l.v-p) )<=0;}DBL PointToLine(Point p, Line l){    return abs((p-l.u)^l.s)/Vlen(l.s);}Point SISC(Line L, Circle C){    Vector ou = L.u-C.o, s=L.s;    DBL a = SQR(s.x) + SQR(s.y);    DBL b = 2*(ou*s);    DBL c = SQR(ou.x) + SQR(ou.y) - SQR(C.r);    DBL d = b*b - 4*a*c;    DBL t = (-b-sqrt(d))/(a+a);    return L.u + s*t;}