Problem Statement |
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| Particles (which can be considered points in 3D-space for the purposes of the problem) can move in an electro-magnetic field. If a particle is charged, its trajectory can be described as spiral, and if it is uncharged, it is just a straight line. Given two particles (one charged and one uncharged) it should be determined whether they can possibly collide or not. Two particles can possibly collide if and only if their trajectories intersect. Some steps have already been made by the physicist to simplify the problem, so the coordinates of the charged particle are represented as follows: x1 = cos(PI * t) y1 = sin(PI * t) z1 = t and for the uncharged particle: x2 = vx * t + x0 y2 = vy * t + y0 z2 = vz * t + z0 Here t is a parameter which can be chosen arbitrarily and independently for both trajectories. Your method will be given 6 integers - vx, vy, vz, x0, y0 and z0, describing the trajectory of the uncharged particle. It should determine whether the two given trajectories intersect or not. If they do, it should return a vector <double> containing exactly 3 elements x, y and z - the coordinates of the point where a collision can happen. If there is more than one such point, it should return a vector <double> containing exactly three zeroes. If collision of the two particles is impossible it should return an empty vector <double>. |
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Definition |
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Notes |
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| - | PI can be considered equal to 3.14159265358979323846. | ||||||||||||
| - | All return values with either an absolute or relative error of less than 1.0E-9 are considered correct. | ||||||||||||
Constraints |
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| - | vx, vy and vz will each be between -10 and 10, inclusive. | ||||||||||||
| - | x0, y0 and z0 will each be between -10 and 10, inclusive. | ||||||||||||
Examples |
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This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.
要做这道题，要注意很多问题：
1.看清题，题目中说了t是independent
2.解一元二次方程要会吧！系数A,B,C的判定不要漏掉
3.特殊情况下A=B=C什么时候有解考虑清楚
这样你就可以过题了！！！
#pragma warning ( disable : 4786 )
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <queue>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
const double PI = acos(-1.0);
#define sz(x) ((int)(x).size())
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define MAXINT 1000000000
#define EPS 1e-8
#define FOR(a,b,c) for(a=(b);(a)<(c);++(a))
#define REP(a,b) FOR(a,0,b)
typedef vector<int> VI;
inline int dblcmp(double a, double b) {
if(fabs(a-b) < EPS) return 0;
return a < b ? -1 : 1;
}
//typedef long long LL;
/*
cos(PI * t1)=vx * t2 + x0
sin(PI * t1)=vy * t2 + y0
t1=vz*t2+z0
*/
class ParticleCollision
{
public:
int vx, vy, vz, x0, y0, z0;
bool check(double t2) {
double x2 = vx * t2 + x0;
double y2 = vy * t2 + y0;
double t1 = vz * t2 + z0;
if(dblcmp(cos(PI * t1), vx * t2 + x0)==0
&& dblcmp(sin(PI * t1),vy * t2 + y0)==0
&& dblcmp(t1,vz*t2+z0)==0)
return true;
return false;
}
vector <double> collision(int _vx, int _vy, int _vz, int _x0, int _y0, int _z0)
{
vz=1;
vx = _vx;
vy = _vy;
vz = _vz;
x0 = _x0;
y0 = _y0;
z0 = _z0;
vector<double> mul(3, 0.0);
vector<double> empty;
int i, j;
double a = vx*vx + vy*vy;
double b = 2 * vx * x0 + 2 * vy * y0;
double c = x0 * x0 + y0 * y0 - 1;
// printf("a = %lf b = %lf c = %lf\n", a, b, c);
double det = b*b-4 * a * c;
if(dblcmp(a, 0.0) == 0) {
if(dblcmp(b, 0.0) == 0) {
if(dblcmp(c, 0.0) == 0) {
if(dblcmp(vz, 0.0) == 0) {
if(dblcmp(x0, cos(PI * z0)) == 0
&& dblcmp(y0, sin(PI * z0)) == 0) {
vector<double> ret;
ret.push_back(x0);
ret.push_back(y0);
ret.push_back(z0);
return ret;
}
}
else return mul;
}
return empty;
}
double t = -c / b;
if(check(t)) {
vector<double> ret;
ret.push_back(vx * t + x0);
ret.push_back(vy * t + y0);
ret.push_back(vz * t + z0);
return ret;
}
return empty;
}
// printf("det = %lf\n", det);
int x = dblcmp(det, 0);
if(x == 1) {
int cnt = 0;
vector<double> ret;
for(i = 0; i < 2; ++i) {
double t;
if(i == 0) t = (-b - sqrt(det)) / 2 / a;
if(i == 1) t = (-b + sqrt(det)) / 2 / a;
if(check(t)) {
ret.push_back(vx * t + x0);
ret.push_back(vy * t + y0);
ret.push_back(vz * t + z0);
cnt++;
}
}
if(cnt == 0) return empty;
else if(cnt == 1) return ret;
else return mul;
}
else if( x== -1) return empty;
else {
double t = (-b + sqrt(det)) / 2 / a;
if(check(t)) {
vector<double> ret;
ret.push_back(vx * t + x0);
ret.push_back(vy * t + y0);
ret.push_back(vz * t + z0);
return ret;
}
return empty;
}
}
};
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