题目链接：点击查看
题目大意：在一个二维平面上，给出一个实心圆柱体，再给出一个可以视为质点的小球，小球初始时位于点 A ，会给出一个速度向量 VA，当小球碰到圆柱体时，会因碰撞弹开，规定碰撞不会丢失动能，问在小球在运动的过程中能不能到达点 B
题目分析：稍微画一下图：
为了方便表示射线，我在无穷远处计算出了一个点 inf，这样射线 A 就转换为了线段 ( A , inf )，同理图中的 l1 和 l3 都是以线段表示的射线，图中点 P 表示为圆心，G1 和 G2 表示 l1 与圆的两个交点，规定 G1 距离点 A 更近，当我们计算出 G1 后，那么直线 ( P , G1 ) 就是法线了，将点 A 根据法线对称得到点 AA，这样就能计算出小球碰撞后的路线了，也就是 l3 所表示的射线，此时只需要分两种情况讨论即可：
- 如果 l1 与圆不相交（相切或相离），只需要判断一下点 B 是否在 l1 上即可
- 如果 l1 与圆相交，求出距离点 A 较近的交点 G1，一顿操作计算出 l2 和 l3 ，判断点 B 是否位于其中一条线段上即可
代码：
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<unordered_map>
#define double long double
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
const int inf=0x3f3f3f3f;
const int N=1e5+100;
// `计算几何模板`
const double eps = 1e-8;
const double pi = acos(-1.0);
const int maxp = 1010;
//`Compares a double to zero`
int sgn(double x){
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
//square of a double
inline double sqr(double x){return x*x;}
/*
* Point
* Point() - Empty constructor
* Point(double _x,double _y) - constructor
* input() - double input
* output() - %.2f output
* operator == - compares x and y
* operator < - compares first by x, then by y
* operator - - return new Point after subtracting curresponging x and y
* operator ^ - cross product of 2d points
* operator * - dot product
* len() - gives length from origin
* len2() - gives square of length from origin
* distance(Point p) - gives distance from p
* operator + Point b - returns new Point after adding curresponging x and y
* operator * double k - returns new Point after multiplieing x and y by k
* operator / double k - returns new Point after divideing x and y by k
* rad(Point a,Point b)- returns the angle of Point a and Point b from this Point
* trunc(double r) - return Point that if truncated the distance from center to r
* rotleft() - returns 90 degree ccw rotated point
* rotright() - returns 90 degree cw rotated point
* rotate(Point p,double angle) - returns Point after rotateing the Point centering at p by angle radian ccw
*/
struct Point{
double x,y;
Point(){}
Point(double _x,double _y){
x = _x;
y = _y;
}
void input(){
cin>>x>>y;
// scanf("%lf%lf",&x,&y);
}
void output(){
printf("%.2f %.2f\n",x,y);
}
bool operator == (Point b)const{
return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
}
bool operator < (Point b)const{
return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x;
}
Point operator -(const Point &b)const{
return Point(x-b.x,y-b.y);
}
//叉积
double operator ^(const Point &b)const{
return x*b.y - y*b.x;
}
//点积
double operator *(const Point &b)const{
return x*b.x + y*b.y;
}
//返回长度
double len(){
return hypot(x,y);//库函数
}
//返回长度的平方
double len2(){
return x*x + y*y;
}
//返回两点的距离
double distance(Point p){
return hypot(x-p.x,y-p.y);
}
Point operator +(const Point &b)const{
return Point(x+b.x,y+b.y);
}
Point operator *(const double &k)const{
return Point(x*k,y*k);
}
Point operator /(const double &k)const{
return Point(x/k,y/k);
}
//`计算pa 和 pb 的夹角`
//`就是求这个点看a,b 所成的夹角`
//`测试 LightOJ1203`
double rad(Point a,Point b){
Point p = *this;
return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));
}
//`化为长度为r的向量`
Point trunc(double r){
double l = len();
if(!sgn(l))return *this;
r /= l;
return Point(x*r,y*r);
}
//`逆时针旋转90度`
Point rotleft(){
return Point(-y,x);
}
//`顺时针旋转90度`
Point rotright(){
return Point(y,-x);
}
//`绕着p点逆时针旋转angle`
Point rotate(Point p,double angle){
Point v = (*this) - p;
double c = cos(angle), s = sin(angle);
return Point(p.x + v.x*c - v.y*s,p.y + v.x*s + v.y*c);
}
};
/*
* Stores two points
* Line() - Empty constructor
* Line(Point _s,Point _e) - Line through _s and _e
* operator == - checks if two points are same
* Line(Point p,double angle) - one end p , another end at angle degree
* Line(double a,double b,double c) - Line of equation ax + by + c = 0
* input() - inputs s and e
* adjust() - orders in such a way that s < e
* length() - distance of se
* angle() - return 0 <= angle < pi
* relation(Point p) - 3 if point is on line
* 1 if point on the left of line
* 2 if point on the right of line
* pointonseg(double p) - return true if point on segment
* parallel(Line v) - return true if they are parallel
* segcrossseg(Line v) - returns 0 if does not intersect
* returns 1 if non-standard intersection
* returns 2 if intersects
* linecrossseg(Line v) - line and seg
* linecrossline(Line v) - 0 if parallel
* 1 if coincides
* 2 if intersects
* crosspoint(Line v) - returns intersection point
* dispointtoline(Point p) - distance from point p to the line
* dispointtoseg(Point p) - distance from p to the segment
* dissegtoseg(Line v) - distance of two segment
* lineprog(Point p) - returns projected point p on se line
* symmetrypoint(Point p) - returns reflection point of p over se
*
*/
struct Line{
Point s,e;
Line(){}
Line(Point _s,Point _e){
s = _s;
e = _e;
}
bool operator ==(Line v){
return (s == v.s)&&(e == v.e);
}
//`根据一个点和倾斜角angle确定直线,0<=angle<pi`
Line(Point p,double angle){
s = p;
if(sgn(angle-pi/2) == 0){
e = (s + Point(0,1));
}
else{
e = (s + Point(1,tan(angle)));
}
}
//ax+by+c=0
Line(double a,double b,double c){
if(sgn(a) == 0){
s = Point(0,-c/b);
e = Point(1,-c/b);
}
else if(sgn(b) == 0){
s = Point(-c/a,0);
e = Point(-c/a,1);
}
else{
s = Point(0,-c/b);
e = Point(1,(-c-a)/b);
}
}
void input(){
s.input();
e.input();
}
void adjust(){
if(e < s)swap(s,e);
}
//求线段长度
double length(){
return s.distance(e);
}
//`返回直线倾斜角 0<=angle<pi`
double angle(){
double k = atan2(e.y-s.y,e.x-s.x);
if(sgn(k) < 0)k += pi;
if(sgn(k-pi) == 0)k -= pi;
return k;
}
//`点和直线关系`
//`1 在左侧`
//`2 在右侧`
//`3 在直线上`
int relation(Point p){
int c = sgn((p-s)^(e-s));
if(c < 0)return 1;
else if(c > 0)return 2;
else return 3;
}
// 点在线段上的判断
bool pointonseg(Point p){
return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;
}
//`两向量平行(对应直线平行或重合)`
bool parallel(Line v){
return sgn((e-s)^(v.e-v.s)) == 0;
}
//`两线段相交判断`
//`2 规范相交`
//`1 非规范相交`
//`0 不相交`
int segcrossseg(Line v){
int d1 = sgn((e-s)^(v.s-s));
int d2 = sgn((e-s)^(v.e-s));
int d3 = sgn((v.e-v.s)^(s-v.s));
int d4 = sgn((v.e-v.s)^(e-v.s));
if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;
return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
(d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
(d4==0 && sgn((e-v.s)*(e-v.e))<=0);
}
//`直线和线段相交判断`
//`-*this line -v seg`
//`2 规范相交`
//`1 非规范相交`
//`0 不相交`
int linecrossseg(Line v){
int d1 = sgn((e-s)^(v.s-s));
int d2 = sgn((e-s)^(v.e-s));
if((d1^d2)==-2) return 2;
return (d1==0||d2==0);
}
//`两直线关系`
//`0 平行`
//`1 重合`
//`2 相交`
int linecrossline(Line v){
if((*this).parallel(v))
return v.relation(s)==3;
return 2;
}
//`求两直线的交点`
//`要保证两直线不平行或重合`
Point crosspoint(Line v){
double a1 = (v.e-v.s)^(s-v.s);
double a2 = (v.e-v.s)^(e-v.s);
return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
}
//点到直线的距离
double dispointtoline(Point p){
return fabs((p-s)^(e-s))/length();
}
//点到线段的距离
double dispointtoseg(Point p){
if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)
return min(p.distance(s),p.distance(e));
return dispointtoline(p);
}
//`返回线段到线段的距离`
//`前提是两线段不相交，相交距离就是0了`
double dissegtoseg(Line v){
return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
}
//`返回点p在直线上的投影`
Point lineprog(Point p){
return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );
}
//`返回点p关于直线的对称点`
Point symmetrypoint(Point p){
Point q = lineprog(p);
return Point(2*q.x-p.x,2*q.y-p.y);
}
};
//圆
struct circle{
Point p;//圆心
double r;//半径
circle(){}
circle(Point _p,double _r){
p = _p;
r = _r;
}
circle(double x,double y,double _r){
p = Point(x,y);
r = _r;
}
//`三角形的外接圆`
//`需要Point的+ / rotate() 以及Line的crosspoint()`
//`利用两条边的中垂线得到圆心`
//`测试：UVA12304`
circle(Point a,Point b,Point c){
Line u = Line((a+b)/2,((a+b)/2)+((b-a).rotleft()));
Line v = Line((b+c)/2,((b+c)/2)+((c-b).rotleft()));
p = u.crosspoint(v);
r = p.distance(a);
}
//`三角形的内切圆`
//`参数bool t没有作用，只是为了和上面外接圆函数区别`
//`测试：UVA12304`
circle(Point a,Point b,Point c,bool t){
Line u,v;
double m = atan2(b.y-a.y,b.x-a.x), n = atan2(c.y-a.y,c.x-a.x);
u.s = a;
u.e = u.s + Point(cos((n+m)/2),sin((n+m)/2));
v.s = b;
m = atan2(a.y-b.y,a.x-b.x) , n = atan2(c.y-b.y,c.x-b.x);
v.e = v.s + Point(cos((n+m)/2),sin((n+m)/2));
p = u.crosspoint(v);
r = Line(a,b).dispointtoseg(p);
}
//输入
void input(){
p.input();
scanf("%lf",&r);
}
//输出
void output(){
printf("%.2lf %.2lf %.2lf\n",p.x,p.y,r);
}
bool operator == (circle v){
return (p==v.p) && sgn(r-v.r)==0;
}
bool operator < (circle v)const{
return ((p<v.p)||((p==v.p)&&sgn(r-v.r)<0));
}
//面积
double area(){
return pi*r*r;
}
//周长
double circumference(){
return 2*pi*r;
}
//`点和圆的关系`
//`0 圆外`
//`1 圆上`
//`2 圆内`
int relation(Point b){
double dst = b.distance(p);
if(sgn(dst-r) < 0)return 2;
else if(sgn(dst-r)==0)return 1;
return 0;
}
//`线段和圆的关系`
//`比较的是圆心到线段的距离和半径的关系`
int relationseg(Line v){
double dst = v.dispointtoseg(p);
if(sgn(dst-r) < 0)return 2;
else if(sgn(dst-r) == 0)return 1;
return 0;
}
//`直线和圆的关系`
//`比较的是圆心到直线的距离和半径的关系`
int relationline(Line v){
double dst = v.dispointtoline(p);
if(sgn(dst-r) < 0)return 2;
else if(sgn(dst-r) == 0)return 1;
return 0;
}
//`两圆的关系`
//`5 相离`
//`4 外切`
//`3 相交`
//`2 内切`
//`1 内含`
//`需要Point的distance`
//`测试：UVA12304`
int relationcircle(circle v){
double d = p.distance(v.p);
if(sgn(d-r-v.r) > 0)return 5;
if(sgn(d-r-v.r) == 0)return 4;
double l = fabs(r-v.r);
if(sgn(d-r-v.r)<0 && sgn(d-l)>0)return 3;
if(sgn(d-l)==0)return 2;
if(sgn(d-l)<0)return 1;
}
//`求两个圆的交点，返回0表示没有交点，返回1是一个交点，2是两个交点`
//`需要relationcircle`
//`测试：UVA12304`
int pointcrosscircle(circle v,Point &p1,Point &p2){
int rel = relationcircle(v);
if(rel == 1 || rel == 5)return 0;
double d = p.distance(v.p);
double l = (d*d+r*r-v.r*v.r)/(2*d);
double h = sqrt(r*r-l*l);
Point tmp = p + (v.p-p).trunc(l);
p1 = tmp + ((v.p-p).rotleft().trunc(h));
p2 = tmp + ((v.p-p).rotright().trunc(h));
if(rel == 2 || rel == 4)
return 1;
return 2;
}
//`求直线和圆的交点，返回交点个数`
int pointcrossline(Line v,Point &p1,Point &p2){
if(!(*this).relationline(v))return 0;
Point a = v.lineprog(p);
double d = v.dispointtoline(p);
d = sqrt(r*r-d*d);
if(sgn(d) == 0){
p1 = a;
p2 = a;
return 1;
}
p1 = a + (v.e-v.s).trunc(d);
p2 = a - (v.e-v.s).trunc(d);
return 2;
}
//`得到过a,b两点，半径为r1的两个圆`
int gercircle(Point a,Point b,double r1,circle &c1,circle &c2){
circle x(a,r1),y(b,r1);
int t = x.pointcrosscircle(y,c1.p,c2.p);
if(!t)return 0;
c1.r = c2.r = r;
return t;
}
//`得到与直线u相切，过点q,半径为r1的圆`
//`测试：UVA12304`
int getcircle(Line u,Point q,double r1,circle &c1,circle &c2){
double dis = u.dispointtoline(q);
if(sgn(dis-r1*2)>0)return 0;
if(sgn(dis) == 0){
c1.p = q + ((u.e-u.s).rotleft().trunc(r1));
c2.p = q + ((u.e-u.s).rotright().trunc(r1));
c1.r = c2.r = r1;
return 2;
}
Line u1 = Line((u.s + (u.e-u.s).rotleft().trunc(r1)),(u.e + (u.e-u.s).rotleft().trunc(r1)));
Line u2 = Line((u.s + (u.e-u.s).rotright().trunc(r1)),(u.e + (u.e-u.s).rotright().trunc(r1)));
circle cc = circle(q,r1);
Point p1,p2;
if(!cc.pointcrossline(u1,p1,p2))cc.pointcrossline(u2,p1,p2);
c1 = circle(p1,r1);
if(p1 == p2){
c2 = c1;
return 1;
}
c2 = circle(p2,r1);
return 2;
}
//`同时与直线u,v相切，半径为r1的圆`
//`测试：UVA12304`
int getcircle(Line u,Line v,double r1,circle &c1,circle &c2,circle &c3,circle &c4){
if(u.parallel(v))return 0;//两直线平行
Line u1 = Line(u.s + (u.e-u.s).rotleft().trunc(r1),u.e + (u.e-u.s).rotleft().trunc(r1));
Line u2 = Line(u.s + (u.e-u.s).rotright().trunc(r1),u.e + (u.e-u.s).rotright().trunc(r1));
Line v1 = Line(v.s + (v.e-v.s).rotleft().trunc(r1),v.e + (v.e-v.s).rotleft().trunc(r1));
Line v2 = Line(v.s + (v.e-v.s).rotright().trunc(r1),v.e + (v.e-v.s).rotright().trunc(r1));
c1.r = c2.r = c3.r = c4.r = r1;
c1.p = u1.crosspoint(v1);
c2.p = u1.crosspoint(v2);
c3.p = u2.crosspoint(v1);
c4.p = u2.crosspoint(v2);
return 4;
}
//`同时与不相交圆cx,cy相切，半径为r1的圆`
//`测试：UVA12304`
int getcircle(circle cx,circle cy,double r1,circle &c1,circle &c2){
circle x(cx.p,r1+cx.r),y(cy.p,r1+cy.r);
int t = x.pointcrosscircle(y,c1.p,c2.p);
if(!t)return 0;
c1.r = c2.r = r1;
return t;
}
//`过一点作圆的切线(先判断点和圆的关系)`
//`测试：UVA12304`
int tangentline(Point q,Line &u,Line &v){
int x = relation(q);
if(x == 2)return 0;
if(x == 1){
u = Line(q,q + (q-p).rotleft());
v = u;
return 1;
}
double d = p.distance(q);
double l = r*r/d;
double h = sqrt(r*r-l*l);
u = Line(q,p + ((q-p).trunc(l) + (q-p).rotleft().trunc(h)));
v = Line(q,p + ((q-p).trunc(l) + (q-p).rotright().trunc(h)));
return 2;
}
//`求两圆相交的面积`
double areacircle(circle v){
int rel = relationcircle(v);
if(rel >= 4)return 0.0;
if(rel <= 2)return min(area(),v.area());
double d = p.distance(v.p);
double hf = (r+v.r+d)/2.0;
double ss = 2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d));
double a1 = acos((r*r+d*d-v.r*v.r)/(2.0*r*d));
a1 = a1*r*r;
double a2 = acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d));
a2 = a2*v.r*v.r;
return a1+a2-ss;
}
//`求圆和三角形pab的相交面积`
//`测试：POJ3675 HDU3982 HDU2892`
double areatriangle(Point a,Point b){
if(sgn((p-a)^(p-b)) == 0)return 0.0;
Point q[5];
int len = 0;
q[len++] = a;
Line l(a,b);
Point p1,p2;
if(pointcrossline(l,q[1],q[2])==2){
if(sgn((a-q[1])*(b-q[1]))<0)q[len++] = q[1];
if(sgn((a-q[2])*(b-q[2]))<0)q[len++] = q[2];
}
q[len++] = b;
if(len == 4 && sgn((q[0]-q[1])*(q[2]-q[1]))>0)swap(q[1],q[2]);
double res = 0;
for(int i = 0;i < len-1;i++){
if(relation(q[i])==0||relation(q[i+1])==0){
double arg = p.rad(q[i],q[i+1]);
res += r*r*arg/2.0;
}
else{
res += fabs((q[i]-p)^(q[i+1]-p))/2.0;
}
}
return res;
}
};
int main()
{
#ifndef ONLINE_JUDGE
// freopen("data.in.txt","r",stdin);
// freopen("data.out.txt","w",stdout);
#endif
// ios::sync_with_stdio(false);
int w;
cin>>w;
int kase=0;
while(w--)
{
bool flag;
Point P,A,B,VA,G1,G2;
double r;
P.input();
cin>>r;
A.input();
VA.input();
B.input();
circle C(P,r);
Line l1(A,A+VA*1e4);
if(C.relationseg(l1)==2)
{
C.pointcrossline(l1,G2,G1);
Point AA=Line(G1,P).symmetrypoint(A);
Line l2(A,G1);
Line l3(G1,G1+((AA-G1)*1e4));
flag=l2.pointonseg(B)||l3.pointonseg(B);
}
else
{
flag=l1.pointonseg(B);
}
printf("Case #%d: %s\n",++kase,flag?"Yes":"No");
}
return 0;
}