Soluție HackerRank pentru Jim Beam, subdomeniul Geometry, în C++14. Include cerința formatată, exemple, explicația pașilor și cod sursă.

  • Problemă: Jim Beam
  • Domeniu: Geometry
  • Limbaj: C++14

Challenge: Jim Beam

Scor cont: 40.0 / 40

Submission status: Accepted

Submission score: 1.0

Submission ID: 464729262

Limbaj: cpp14

Link challenge: https://www.hackerrank.com/challenges/jim-beam/problem

Cerință

Jim likes to play with laser beams.
Jim stays at point (0,0).
There is a mirror at point (X_m, Y_m) and a wall between points (X_1, Y_1) and (X_2, Y_2).

Jim wants to find out if he can point the laser beam on the mirror.

Input Format
First line contains the number of test cases, T.
Each subsequent line contains one test case: space separated integers that denote endpoints of the wall X_1, Y_1, X_2, Y_2 and position of the mirror X_m, Y_m.

Output Format
The answer for each test case: Display `YES` if Jim can point the laser beam to the mirror, otherwise display `NO` .

Constraints
1 ≤ T ≤ 100
0 ≤ |X_1|,|Y_1|,|X_2|,|Y_2|,|X_m|,|Y_m| ≤ 10^6
Mirror doesn't have common points with wall.
Wall doesn't pass through the point (0,0).

Sample Input

5
1 2 2 1 2 2
-1 1 1 1 1 -1
1 1 2 2 3 3
2 2 3 3 1 1
0 1 1 1 0 2

Sample Output

NO
YES
NO
YES
NO

Cod sursă

//jim_beam.cpp
//Jim Beam
//Ad Infinitum - Math Programming Contest August'14
//Author: derekhh

#include<iostream>
#include<algorithm>
using namespace std;

#define EPS 1e-8
#define geq(x,y) ((x)+EPS>=(y))

struct Point
{
	double x, y;
	Point(double x0 = 0, double y0 = 0) :x(x0), y(y0){}
};

struct Line
{
	Point p1, p2;
};

double times(Point p0, Point p1, Point p2)
{
	return (p1.x - p0.x)*(p2.y - p0.y) - (p1.y - p0.y)*(p2.x - p0.x);
}

bool LineSegIntersect(Line L1, Line L2)
{
	return(geq(max(L1.p1.x, L1.p2.x), min(L2.p1.x, L2.p2.x))
		&& geq(max(L2.p1.x, L2.p2.x), min(L1.p1.x, L1.p2.x))
		&& geq(max(L1.p1.y, L1.p2.y), min(L2.p1.y, L2.p2.y))
		&& geq(max(L2.p1.y, L2.p2.y), min(L1.p1.y, L1.p2.y))
		&& times(L1.p1, L2.p1, L1.p2)*times(L1.p1, L2.p2, L1.p2) <= EPS
		&&times(L2.p1, L1.p1, L2.p2)*times(L2.p1, L1.p2, L2.p2) <= EPS);
}

int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		Line L1, L2;
		L1.p1 = Point(0, 0);
		cin >> L2.p1.x >> L2.p1.y >> L2.p2.x >> L2.p2.y >> L1.p2.x >> L1.p2.y;
		if (LineSegIntersect(L1, L2))
			cout << "NO" << endl;
		else
			cout << "YES" << endl;
	}
	return 0;
}
HackerRank Geometry – Jim Beam