hdu5572An Easy Physics Problem+计算几何

An Easy Physics Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1501 Accepted Submission(s): 289

Problem Description
On an infinite smooth table, there’s a big round fixed cylinder and a little ball whose volume can be ignored.

Currently the ball stands still at point A, then we’ll give it an initial speed and a direction. If the ball hits the cylinder, it will bounce back with no energy losses.

We’re just curious about whether the ball will pass point B after some time.

Input
First line contains an integer T, which indicates the number of test cases.

Every test case contains three lines.

The first line contains three integers Ox, Oy and r, indicating the center of cylinder is (Ox,Oy) and its radius is r.

The second line contains four integers Ax, Ay, Vx and Vy, indicating the coordinate of A is (Ax,Ay) and the initial direction vector is (Vx,Vy).

The last line contains two integers Bx and By, indicating the coordinate of point B is (Bx,By).

⋅ 1 ≤ T ≤ 100.

⋅ |Ox|,|Oy|≤ 1000.

⋅ 1 ≤ r ≤ 100.

⋅ |Ax|,|Ay|,|Bx|,|By|≤ 1000.

⋅ |Vx|,|Vy|≤ 1000.

⋅ Vx≠0 or Vy≠0.

⋅ both A and B are outside of the cylinder and they are not at same position.

Output
For every test case, you should output “Case #x: y”, where x indicates the case number and counts from 1. y is “Yes” if the ball will pass point B after some time, otherwise y is “No”.

Sample Input

2
0 0 1
2 2 0 1
-1 -1
0 0 1
-1 2 1 -1
1 2

Sample Output

Case #1: No
Case #2: Yes

Source
2015ACM/ICPC亚洲区上海站-重现赛（感谢华东理工）

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题意：一个求从A点除法给出方向，问能不能到达B点。如果点碰到小球，A会完全碰撞。。
其实也不是很难：但是啊：
思路，首先判断我们可以有方向构造一个temp点，这样A，temp来表示A的轨迹，
（1）然后判断这天直线与园是否相交，如果不想交或相切，判断A能否经过B。
（2）如果有两个交点，找到那个与A最近的那个L，《1》判断B是否在A,L中间，以及L,A射线外。《2》，找到A关于L，O (圆心，完全碰撞）的对称点B2。然后判断点从L出发到B2,能否经过B点。。
判断B是否在A->temp上。。就是要差积为0（两直线围的四边形面积）（平行），点积大于0（点积>0同向，=0垂直,<0反向）。判断点C在线段AB中间，只需要差积大于0，点积小于0.

思路就这样。但是起码wa了6发。不知所错啊。。经过反复的对拍。。发现找对称点的那个函数有问题（换个模板就能过。。目测是数据溢出？long double 啊。。算了。换模板吧。。）

#include <bits/stdc++.h>using namespace std;#define maxn 100005const double eps = 1e-5;const double INF = 1e20;const double pi = acos (-1.0);int dcmp (double x) {    if (fabs (x) < eps) return 0;    return (x < 0 ? -1 : 1);}//*************点struct Point {    double x, y;    Point (double _x = 0, double _y = 0):x(_x), y(_y) {}    bool operator < (const Point &b) const {        return (dcmp (x-b.x) == 0 ? dcmp (y-b.y) < 0 : x < b.x);    }    Point operator + (const Point &b) const {        return Point (x+b.x, y+b.y);    }    Point operator - (const Point &b) const {        return Point (x-b.x, y-b.y);    }    Point operator * (double a) {        return Point (x*a, y*a);    }    Point operator / (double a) {        return Point (x/a, y/a);    }    double len2 () {//返回长度的平方        return x*x + y*y;    }    double len () {//返回长度        return sqrt (len2 ());    }    Point change_len (double r) {//转化为长度为r的向量        double l = len ();        if (dcmp (l) == 0) return *this;//零向量返回自身        r /= l;        return Point (x*r, y*r);    }};//计算两个向量的叉积long double cross(const Point &a,const Point &b){    return a.x*b.y-a.y*b.x;}//计算两个点的点积long double dot(const Point &a,const Point &b){    return a.x*b.x+a.y*b.y;}long double xmult(Point p0,Point p1,Point p2) {    return cross((p1-p0),(p2-p0));}//dot(p1-p0).(p2-p0)long double dmult(Point p0,Point p1,Point p2){    return (p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y);}long double dis (Point a, Point b) {//两个点的距离    Point p = b-a; return p.len ();}//************直线 线段struct Line {    Point s, e;//直线的两个点    double k;//极角    Line () {}    Line (Point _s, Point _e) {        s = _s, e = _e;        k = atan2 (e.y - s.y,e.x - s.x);    }    double length () {//求线段长度        return dis (s, e);    }};//点和直线的关系int relation (Point p, Line l) {    //1:在左侧 2:在右侧 3:在直线上    int c = dcmp (cross (p-l.s, l.e-l.s));    if (c < 0) return 1;    else if (c > 0) return 2;    else return 3;}//判断点在线段上bool point_on_seg (Point p, Line l) {    return dcmp (cross (p-l.s, l.e-l.s)) == 0 &&    dcmp (dot (p-l.s, p-l.e) <= 0);    //如果忽略端点交点改成小于号就好了}//判断点在射线上bool point_on_halfline (Point p, Line l) {    int id = relation (p, l);    if (id != 3) return 0;    return dcmp (dot (p-l.s, l.e-l.s)) >= 0;}//点到直线的距离double point_to_line (Point p, Line a) {    return fabs (cross (p-a.s, a.e-a.s) / a.length ());}//点在直线上的投影Point projection (Point p, Line a) {    return a.s + (((a.e-a.s) * dot (a.e-a.s, p-a.s)) / (a.e-a.s).len2() );}//点关于直线的对称点Point symmetry (Point p, Line a) {    Point q = projection (p, a);    return Point (2*q.x-p.x, 2*q.y-p.y);}//***************圆struct Circle {    //圆心 半径    Point p;    double r;    Circle () {}    Circle (Point _p, double _r) : p(_p), r(_r) {}};//直线和圆的关系int relation (Line a, Circle b) {    //0:相离 1:相切 2:相交    double p = point_to_line (b.p, a);    if (dcmp (p-b.r) == 0) return 1;    return (dcmp (p-b.r) < 0 ? 2 : 0);}//直线和圆的交点int line_circle_intersection (Line v, Circle u, Point &p1, Point &p2) {    //返回交点个数 交点保存在引用中    if (!relation (v, u)) return 0;    Point a = projection (u.p, v);    double d = point_to_line (u.p, v);    d = sqrt (u.r*u.r - d*d);    if (dcmp (d) == 0) {        p1 = a, p2 = a;        return 1;    }    p1 = a + (v.e-v.s).change_len (d);    p2 = a - (v.e-v.s).change_len (d);    return 2;}Circle c;Point A,B;long double v1,v2;int main(){    int t;    cin>>t;    for(int cas=1;cas<=t;cas++){        cin>>c.p.x>>c.p.y>>c.r;        cin>>A.x>>A.y>>v1>>v2;        cin>>B.x>>B.y;        cout<<"Case #"<<cas<<": ";        Point temp=Point(A.x+v1,A.y+v2);        int num;        Point p1,p2;        num=line_circle_intersection(Line(A,temp),c,p1,p2);        if(num<2){            if(dmult(A,temp,B)>eps&&fabs(xmult(A,temp,B))<eps) cout<<"Yes"<<endl;            //if(point_on_halfline(B,Line(A,temp))) cout<<"Yes"<<endl;            else cout<<"No"<<endl;        }        else{       //相交            Point L;            if(dis(A,p1)>dis(A,p2)) L=p2;            else L=p1;            //if(point_on_seg(B,Line(A,L))){            if(abs(xmult(B,A,L))<eps&&dmult(B,A,L)<-eps){                cout<<"Yes"<<endl;                continue;            }            Point B2=symmetry(A,Line(L,c.p));            //if(point_on_halfline(B,Line(L,B2))) cout<<"Yes"<<endl;            if(abs(xmult(L,B2,B))<eps&&dmult(L,B2,B)>eps) cout<<"Yes"<<endl;            else cout<<"No"<<endl;        }    }    return 0;}