An Easy Physics Problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3840 Accepted Submission(s): 765

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亞洲區上海站-重現賽（感謝華東理工）

先判斷射線和圓交點個數，如果小於2再看是否B在A的前進方向上，沒有則NO，否則YES。如果等於2，就先找到第一個交點，將這個交點和圓心連成直線，那麼A的路徑關於這條直線對稱，那麼如果A關於此直線的對稱點在圓心->B路徑上，則可以相撞，否則不行。 這裡有一個小問題，如果反過來求B關於此直線的對稱點在圓心->A路徑上，是會WA的。

#include <iostream>
#include <cstdio>
#include <cstring>
#include<algorithm>
#include <cstdlib>
#include <cmath>
using namespace std;
const double eps = 1e-8;
int sgn(double x) {
	if (fabs(x) < eps)return 0;
	if (x < 0)return -1;
	else return 1;
}
struct point {
	double x, y;
	point() {}
	point(double x, double y) : x(x), y(y) {}
	void input() {
		scanf("%lf%lf", &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 {		//返回減去後的新點
		return point(x - b.x, y - b.y);
	}
	point operator +(const point &b)const {		//返回加上後的新點
		return point(x + b.x, y + b.y);
	}
	point operator *(const double &k)const {	//返回相乘後的新點
		return point(x * k, y * k);
	}
	point operator /(const double &k)const {	//返回相除後的新點
		return point(x / k, y / k);
	}
	double operator ^(const point &b)const {	//叉乘
		return x*b.y - y*b.x;
	}
	double operator *(const point &b)const {	//點乘
		return x*b.x + y*b.y;
	}
	double len() {		//返回長度
		return hypot(x, y);
	}
	double len2() {		//返回長度的平方
		return x*x + y*y;
	}
	point trunc(double r) {
		double l = len();
		if (!sgn(l))return *this;
		r /= l;
		return point(x*r, y*r);
	}
};
struct line {
	point s;
	point e;
	line() {

	}
	line(point _s, point _e) {
		s = _s;
		e = _e;
	}
	bool operator ==(line v) {
		return (s == v.s) && (e == v.e);
	}
	//返回點p在直線上的投影
	point lineprog(point p) {
		return s + (((e - s)*((e - s)*(p - s))) / ((e - s).len2()));
	}
	//返回點p關於直線的對稱點
	point symmetrypoint(point p) {
		point q = lineprog(p);
		return point(2 * q.x - p.x, 2 * q.y - p.y);
	}
	//點是否線上段上
	bool pointonseg(point p) {
		return sgn((p - s) ^ (e - s)) == 0 && sgn((p - s)*(p - e)) <= 0;
	}
};
struct circle {//圓
	double r;	//半徑
	point p;	//圓心
	void input() {
		p.input();
		scanf("%lf", &r);
	}
	circle() { }
	circle(point _p, double _r) {
		p = _p;
		r = _r;
	}
	circle(double x, double y, double _r) {
		p = point(x, y);
		r = _r;
	}
	//求直線和圓的交點，返回交點個數
	int pointcrossline(line l, point &r1, point &r2) {
		double dx = l.e.x - l.s.x, dy = l.e.y - l.s.y;
		double A = dx*dx + dy*dy;
		double B = 2 * dx * (l.s.x - p.x) + 2 * dy * (l.s.y - p.y);
		double C = (l.s.x - p.x)*(l.s.x - p.x) + (l.s.y - p.y)*(l.s.y - p.y) - r*r;
		double del = B*B - 4 * A * C;
		if (sgn(del) < 0)  return 0;
		int cnt = 0;
		double t1 = (-B - sqrt(del)) / (2 * A);
		double t2 = (-B + sqrt(del)) / (2 * A);
		if (sgn(t1) >= 0) {
			r1 = point(l.s.x + t1 * dx, l.s.y + t1 * dy);
			cnt++;
		}
		if (sgn(t2) >= 0) {
			r2 = point(l.s.x + t2 * dx, l.s.y + t2 * dy);
			cnt++;
		}
		return cnt;
	}
};
point A, V, B;
circle tc;
point r1, r2;
int main() {
	int t, d = 1;
	scanf("%d", &t);
	while (t--) {
		tc.input();
		A.input();
		V.input();
		B.input();
		int f = 0;
		int num = tc.pointcrossline(line(A, A + V), r1, r2);
		if (num < 2) {
			point t = B - A;
			if (t.trunc(1) == V.trunc(1)) f = 1;
			else f = 0;
		}
		else {
			line l = line(tc.p, r1);
			line l1 = line(A, r1);
			line l2 = line(r1, B);
			point t = l.symmetrypoint(A);
			if (l1.pointonseg(B))f = 1;
			else if (l2.pointonseg(t))f = 1;		//求B的對稱點會WA
			else f = 0;
		}
		if (f == 1)
			printf("Case #%d: Yes\n", d++);
		else
			printf("Case #%d: No\n", d++);
	}
	return 0;
}