HDU 4793 Collision (2013长沙现场赛，简单计算几何)

Collision

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 685    Accepted Submission(s): 248
Special Judge

Problem Description

There's a round medal fixed on an ideal smooth table, Fancy is trying to throw some coins and make them slip towards the medal to collide. There's also a round range which shares exact the same center as the round medal, and radius of the medal is strictly less than radius of the round range. Since that the round medal is fixed and the coin is a piece of solid metal, we can assume that energy of the coin will not lose, the coin will collide and then moving as reflect.
Now assume that the center of the round medal and the round range is origin ( Namely (0, 0) ) and the coin's initial position is strictly outside the round range.
Given radius of the medal R_m, radius of coin r, radius of the round range R, initial position (x, y) and initial speed vector (vx, vy) of the coin, please calculate the total time that any part of the coin is inside the round range. Please note that the coin might not even touch the medal or slip through the round range.

 

Input

There will be several test cases. Each test case contains 7 integers R_m, R, r, x, y, vx and vy in one line. Here 1 ≤ R_m < R ≤ 2000, 1 ≤ r ≤ 1000, R + r < |(x, y)| ≤ 20000, 1 ≤ |(vx, vy)| ≤ 100.

 

Output

For each test case, please calculate the total time that any part of the coin is inside the round range. Please output the time in one line, an absolute error not more than 1e -3 is acceptable.

 

Sample Input

5 20 1 0 100 0 -1 5 20 1 30 15 -1 0

 

Sample Output

30.000 29.394

 

Source

2013 Asia Changsha Regional Contest

 

 

【题意】：

圆心在原点的两个同心圆，半径是Rm和R，还有一个半径为r的硬币，给出初始位置和速度矢量，

硬币与内圆碰撞后会等动能反弹，求硬币在大圆区域内运动的时间。。。

 

【解题思路】：

在t时刻，硬币的位置为P(x+Vx*t, y+Vy*t)把P分别代到x^2+y^2=(R+r)^2 和 x^2+y^2=(Rm+r)^2（一定要加 r ）中，可以求出硬币与大小圆接触的时间点

根据方程的解的情况判断：若与大圆无2个交点输出0；若与内圆无2个交点输出t1-t2；

若发生碰撞，则输出2*(t3-t1); —— 若碰撞，则入射与反射路径对称，长度必然相等

WA了一次，没有判断方程解的非负性

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define eps 1e-8    /* >0:>eps--- <0:<-eps */
#define zero(x)(((x)>0?(x):-(x))<eps)
#define PI acos(-1.0)
#define LL long long
#define maxn 100100
#define IN freopen("in.txt","r",stdin);
using namespace std;

int sign(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0? -1:1;
}

double Rm,R,r,x,y,vx,vy;
double v;

int main(int argc, char const *argv[])
{
    //IN;

    while(scanf("%lf%lf%lf%lf%lf%lf%lf",&Rm,&R,&r,&x,&y,&vx,&vy)!=EOF)
    {
        double a,b,c,del;
        a = vx*vx+vy*vy;
        b = 2.0*x*vx + 2.0*y*vy;
        c = x*x + y*y - (R+r)*(R+r);

        del = (b*b - 4.0*a*c);
        if(sign(del)<=0) {printf("0
");continue;}

        del=sqrt(del);
        double t1,t2;
        t1 = (-b+del)/(2.0*a);t2 = (-b-del)/(2.0*a);
        if(t1>t2) swap(t1,t2);
        if(t1<0) {printf("0
");continue;}

        a = vx*vx+vy*vy;
        b = 2.0*x*vx + 2.0*y*vy;
        c = x*x + y*y - (Rm+r)*(Rm+r);

        del = (b*b - 4.0*a*c);
        if(sign(del)<=0) {printf("%.3lf
",fabs(t1-t2));continue;}

        del=sqrt(del);
        double t3,t4;
        t3 = (-b+del)/(2.0*a);t4 = (-b-del)/(2.0*a);
        if(t3>t4) swap(t3,t4);
        if(t3<0) {printf("%.3lf
",fabs(t1-t2));continue;}

        printf("%.3lf
",2.0*fabs(t3-t1));
    }

    return 0;
}