Ņemot vērā N punktus 2-dimensiju telpā, mums jāatrod trīs punkti tā, lai trijstūrī, kas izveidots, izvēloties šos punktus, iekšā nebūtu citu punktu. Visi dotie punkti neatradīsies uz vienas taisnes, tāpēc risinājums vienmēr pastāvēs.
Piemēri:
In above diagram possible triangle with no point
inside can be formed by choosing these triplets
[(0 0) (2 0) (1 1)]
[(0 0) (1 1) (0 2)]
[(1 1) (2 0) (2 2)]
[(1 1) (0 2) (2 2)]
So any of the above triplets can be the final answer.
Risinājums ir balstīts uz faktu, ka, ja eksistē trijstūris(-i) bez punktiem, tad mēs varam izveidot trīsstūri ar jebkuru nejaušu punktu starp visiem punktiem.
Mēs varam atrisināt šo problēmu, pārmeklējot visus trīs punktus pa vienam. Pirmo punktu var izvēlēties nejauši. Pēc pirmā punkta izvēles mums ir nepieciešami divi punkti, lai to slīpums būtu atšķirīgs un neviens punkts neatrastos šo trīs punktu trīsstūrī. Mēs to varam izdarīt, izvēloties otro punktu kā tuvāko punktu pirmajam un trešo punktu kā otro tuvāko punktu ar atšķirīgu slīpumu. Lai to izdarītu, vispirms atkārtojam visus punktus un izvēlamies punktu, kas ir vistuvāk pirmajam, un apzīmējam to kā vajadzīgā trīsstūra otro punktu. Pēc tam mēs atkārtojam vēl vienu reizi, lai atrastu punktu, kuram ir atšķirīgs slīpums un mazākais attālums, kas būs mūsu trīsstūra trešais punkts.
foreach mašīnrakstāC++
// C/C++ program to find triangle with no point inside #include using namespace std; // method to get square of distance between // (x1 y1) and (x2 y2) int getDistance(int x1 int y1 int x2 int y2) { return (x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1); } // Method prints points which make triangle with no // point inside void triangleWithNoPointInside(int points[][2] int N) { // any point can be chosen as first point of triangle int first = 0; int second third; int minD = INT_MAX; // choose nearest point as second point of triangle for (int i = 0; i < N; i++) { if (i == first) continue; // Get distance from first point and choose // nearest one int d = getDistance(points[i][0] points[i][1] points[first][0] points[first][1]); if (minD > d) { minD = d; second = i; } } // Pick third point by finding the second closest // point with different slope. minD = INT_MAX; for (int i = 0; i < N; i++) { // if already chosen point then skip them if (i == first || i == second) continue; // get distance from first point int d = getDistance(points[i][0] points[i][1] points[first][0] points[first][1]); /* the slope of the third point with the first point should not be equal to the slope of second point with first point (otherwise they'll be collinear) and among all such points we choose point with the smallest distance */ // here cross multiplication is compared instead // of division comparison if ((points[i][0] - points[first][0]) * (points[second][1] - points[first][1]) != (points[second][0] - points[first][0]) * (points[i][1] - points[first][1]) && minD > d) { minD = d; third = i; } } cout << points[first][0] << ' ' << points[first][1] << endl; cout << points[second][0] << ' ' << points[second][1] << endl; cout << points[third][0] << ' ' << points[third][1] << endl; } // Driver code to test above methods int main() { int points[][2] = {{0 0} {0 2} {2 0} {2 2} {1 1}}; int N = sizeof(points) / sizeof(points[0]); triangleWithNoPointInside(points N); return 0; }
Java // Java program to find triangle // with no point inside import java.io.*; class GFG { // method to get square of distance between // (x1 y1) and (x2 y2) static int getDistance(int x1 int y1 int x2 int y2) { return (x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1); } // Method prints points which make triangle with no // point inside static void triangleWithNoPointInside(int points[][] int N) { // any point can be chosen as first point of triangle int first = 0; int second =0; int third =0; int minD = Integer.MAX_VALUE; // choose nearest point as second point of triangle for (int i = 0; i < N; i++) { if (i == first) continue; // Get distance from first point and choose // nearest one int d = getDistance(points[i][0] points[i][1] points[first][0] points[first][1]); if (minD > d) { minD = d; second = i; } } // Pick third point by finding the second closest // point with different slope. minD = Integer.MAX_VALUE; for (int i = 0; i < N; i++) { // if already chosen point then skip them if (i == first || i == second) continue; // get distance from first point int d = getDistance(points[i][0] points[i][1] points[first][0] points[first][1]); /* the slope of the third point with the first point should not be equal to the slope of second point with first point (otherwise they'll be collinear) and among all such points we choose point with the smallest distance */ // here cross multiplication is compared instead // of division comparison if ((points[i][0] - points[first][0]) * (points[second][1] - points[first][1]) != (points[second][0] - points[first][0]) * (points[i][1] - points[first][1]) && minD > d) { minD = d; third = i; } } System.out.println(points[first][0] + ' ' + points[first][1]); System.out.println(points[second][0]+ ' ' + points[second][1]) ; System.out.println(points[third][0] +' ' + points[third][1]); } // Driver code public static void main (String[] args) { int points[][] = {{0 0} {0 2} {2 0} {2 2} {1 1}}; int N = points.length; triangleWithNoPointInside(points N); } } // This article is contributed by vt_m.
Python 3 # Python3 program to find triangle # with no point inside import sys # method to get square of distance between # (x1 y1) and (x2 y2) def getDistance(x1 y1 x2 y2): return (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1) # Method prints points which make triangle # with no point inside def triangleWithNoPointInside(points N): # any point can be chosen as # first point of triangle first = 0 second = 0 third = 0 minD = sys.maxsize # choose nearest point as # second point of triangle for i in range(0 N): if i == first: continue # Get distance from first point and choose # nearest one d = getDistance(points[i][0] points[i][1] points[first][0] points[first][1]) if minD > d: minD = d second = i # Pick third point by finding the second closest # point with different slope. minD = sys.maxsize for i in range (0 N): # if already chosen point then skip them if i == first or i == second: continue # get distance from first point d = getDistance(points[i][0] points[i][1] points[first][0] points[first][1]) ''' the slope of the third point with the first point should not be equal to the slope of second point with first point (otherwise they'll be collinear) and among all such points we choose point with the smallest distance ''' # here cross multiplication is compared instead # of division comparison if ((points[i][0] - points[first][0]) * (points[second][1] - points[first][1]) != (points[second][0] - points[first][0]) * (points[i][1] - points[first][1]) and minD > d) : minD = d third = i print(points[first][0] ' ' points[first][1]) print(points[second][0] ' ' points[second][1]) print(points[third][0] ' ' points[third][1]) # Driver code points = [[0 0] [0 2] [2 0] [2 2] [1 1]] N = len(points) triangleWithNoPointInside(points N) # This code is contributed by Gowtham Yuvaraj
C# using System; // C# program to find triangle // with no point inside public class GFG { // method to get square of distance between // (x1 y1) and (x2 y2) public static int getDistance(int x1 int y1 int x2 int y2) { return (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1); } // Method prints points which make triangle with no // point inside public static void triangleWithNoPointInside(int[][] points int N) { // any point can be chosen as first point of triangle int first = 0; int second = 0; int third = 0; int minD = int.MaxValue; // choose nearest point as second point of triangle for (int i = 0; i < N; i++) { if (i == first) { continue; } // Get distance from first point and choose // nearest one int d = getDistance(points[i][0] points[i][1] points[first][0] points[first][1]); if (minD > d) { minD = d; second = i; } } // Pick third point by finding the second closest // point with different slope. minD = int.MaxValue; for (int i = 0; i < N; i++) { // if already chosen point then skip them if (i == first || i == second) { continue; } // get distance from first point int d = getDistance(points[i][0] points[i][1] points[first][0] points[first][1]); /* the slope of the third point with the first point should not be equal to the slope of second point with first point (otherwise they'll be collinear) and among all such points we choose point with the smallest distance */ // here cross multiplication is compared instead // of division comparison if ((points[i][0] - points[first][0]) * (points[second][1] - points[first][1]) != (points[second][0] - points[first][0]) * (points[i][1] - points[first][1]) && minD > d) { minD = d; third = i; } } Console.WriteLine(points[first][0] + ' ' + points[first][1]); Console.WriteLine(points[second][0] + ' ' + points[second][1]); Console.WriteLine(points[third][0] + ' ' + points[third][1]); } // Driver code public static void Main(string[] args) { int[][] points = new int[][] { new int[] {0 0} new int[] {0 2} new int[] {2 0} new int[] {2 2} new int[] {1 1} }; int N = points.Length; triangleWithNoPointInside(points N); } } // This code is contributed by Shrikant13
JavaScript <script> // javascript program to find triangle // with no point inside // method to get square of distance between // (x1 y1) and (x2 y2) function getDistance(x1 y1 x2 y2) { return (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1); } // Method prints points which make triangle with no // point inside function triangleWithNoPointInside(points N) { // any point can be chosen as first point of triangle var first = 0; var second = 0; var third = 0; var minD = Number.MAX_VALUE; // choose nearest point as second point of triangle for (i = 0; i < N; i++) { if (i == first) continue; // Get distance from first point and choose // nearest one var d = getDistance(points[i][0] points[i][1] points[first][0] points[first][1]); if (minD > d) { minD = d; second = i; } } // Pick third point by finding the second closest // point with different slope. minD = Number.MAX_VALUE; for (i = 0; i < N; i++) { // if already chosen point then skip them if (i == first || i == second) continue; // get distance from first point var d = getDistance(points[i][0] points[i][1] points[first][0] points[first][1]); /* * the slope of the third point with the first point should not be equal to the * slope of second point with first point (otherwise they'll be collinear) and * among all such points we choose point with the smallest distance */ // here cross multiplication is compared instead // of division comparison if ((points[i][0] - points[first][0]) * (points[second][1] - points[first][1]) != (points[second][0] - points[first][0]) * (points[i][1] - points[first][1]) && minD > d) { minD = d; third = i; } } document.write(points[first][0] + ' ' + points[first][1]+'
'); document.write(points[second][0] + ' ' + points[second][1]+'
'); document.write(points[third][0] + ' ' + points[third][1]+'
'); } // Driver code var points = [ [ 0 0 ] [ 0 2 ] [ 2 0 ] [ 2 2 ] [ 1 1 ] ]; var N = points.length; triangleWithNoPointInside(points N); // This code contributed by umadevi9616 </script>
Izvade:
0 0
1 1
0 2
Laika sarežģītība: O(n)
Palīgtelpa: O(1)
Šī raksta autors ir Utkarsh Trivedi .
2. pieeja: brutāla spēka izmantošana
Šis kods atkārto visus iespējamos trīsstūrus, kurus var izveidot no dotās punktu kopas, un pārbauda, vai katrā trijstūrī atrodas kāds cits punkts. Ja tiek atrasts trīsstūris, kurā nav neviena punkta, kods atgriež trīsstūri. Pretējā gadījumā tas atgriež Nav.
Algoritms
1. Atkārtojiet visus iespējamos trīsstūrus ar virsotnēm no dotajiem punktiem.
2. Katram trīsstūrim pārbaudiet, vai kāds no atlikušajiem punktiem atrodas trijstūrī.
3. Ja neviena trijstūra iekšpusē neatrodas neviens punkts, atgriež pirmā atrastā trijstūra koordinātas.
#include #include #include using namespace std; // Function to calculate the area of a triangle given its three vertices double area(double x1 double y1 double x2 double y2 double x3 double y3) { return abs((x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) / 2.0); } // Function to check if a point (x y) is inside a triangle defined by its vertices (x1 y1) (x2 y2) and (x3 y3) bool isInsideTriangle(double x1 double y1 double x2 double y2 double x3 double y3 double x double y) { double A = area(x1 y1 x2 y2 x3 y3); double A1 = area(x y x2 y2 x3 y3); double A2 = area(x1 y1 x y x3 y3); double A3 = area(x1 y1 x2 y2 x y); return abs(A - (A1 + A2 + A3)) < 1e-9; // Use a small epsilon value for comparison due to floating-point precision } // Function to find three points from a list that do not form a triangle with any other point inside vector<vector<double>> noPointInsideTriangle(vector<vector<double>> points) { for (int i = 0; i < points.size(); ++i) { for (int j = i + 1; j < points.size(); ++j) { for (int k = j + 1; k < points.size(); ++k) { bool inside = false; for (int l = 0; l < points.size(); ++l) { if (l != i && l != j && l != k) { if (isInsideTriangle(points[i][0] points[i][1] points[j][0] points[j][1] points[k][0] points[k][1] points[l][0] points[l][1])) { inside = true; break; } } } if (!inside) { vector<vector<double>> result = {points[i] points[j] points[k]}; return result; } } } } return vector<vector<double>>(); // Return an empty vector if no such set of points is found } int main() { vector<vector<double>> points = {{0 0} {0 2} {2 0} {2 2} {1 1}}; vector<vector<double>> result = noPointInsideTriangle(points); if (!result.empty()) { cout << 'Points that do not form a triangle with any other point inside:' << endl; for (const auto& point : result) { cout << '(' << point[0] << ' ' << point[1] << ')' << endl; } } else { cout << 'No such set of points found.' << endl; } return 0; }
Java import java.util.ArrayList; import java.util.List; public class Main { static double area(int x1 int y1 int x2 int y2 int x3 int y3) { return Math.abs((x1*(y2-y3) + x2*(y3-y1) + x3*(y1-y2))/2.0); } static boolean isInsideTriangle(int x1 int y1 int x2 int y2 int x3 int y3 int x int y) { double A = area(x1 y1 x2 y2 x3 y3); double A1 = area(x y x2 y2 x3 y3); double A2 = area(x1 y1 x y x3 y3); double A3 = area(x1 y1 x2 y2 x y); return A == A1 + A2 + A3; } static List<List<Integer>> noPointInsideTriangle(List<List<Integer>> points) { for (int i = 0; i < points.size(); i++) { for (int j = i+1; j < points.size(); j++) { for (int k = j+1; k < points.size(); k++) { boolean inside = false; for (int l = 0; l < points.size(); l++) { if (l != i && l != j && l != k) { if (isInsideTriangle(points.get(i).get(0) points.get(i).get(1) points.get(j).get(0) points.get(j).get(1) points.get(k).get(0) points.get(k).get(1) points.get(l).get(0) points.get(l).get(1))) { inside = true; break; } } } if (!inside) { List<List<Integer>> result = new ArrayList<>(); result.add(points.get(i)); result.add(points.get(j)); result.add(points.get(k)); return result; } } } } return null; } public static void main(String[] args) { List<List<Integer>> points = new ArrayList<>(); points.add(new ArrayList<Integer>(){{add(0); add(0);}}); points.add(new ArrayList<Integer>(){{add(0); add(2);}}); points.add(new ArrayList<Integer>(){{add(2); add(0);}}); points.add(new ArrayList<Integer>(){{add(2); add(2);}}); points.add(new ArrayList<Integer>(){{add(1); add(1);}}); List<List<Integer>> result = noPointInsideTriangle(points); if (result != null) { System.out.println(result); } else { System.out.println('No triangle found.'); } } }
Python3 def area(x1 y1 x2 y2 x3 y3): return abs((x1*(y2-y3) + x2*(y3-y1) + x3*(y1-y2))/2.0) def is_inside_triangle(x1 y1 x2 y2 x3 y3 x y): A = area(x1 y1 x2 y2 x3 y3) A1 = area(x y x2 y2 x3 y3) A2 = area(x1 y1 x y x3 y3) A3 = area(x1 y1 x2 y2 x y) return A == A1 + A2 + A3 def no_point_inside_triangle(points): for i in range(len(points)): for j in range(i+1 len(points)): for k in range(j+1 len(points)): inside = False for l in range(len(points)): if l != i and l != j and l != k: if is_inside_triangle(points[i][0] points[i][1] points[j][0] points[j][1] points[k][0] points[k][1] points[l][0] points[l][1]): inside = True break if not inside: return [points[i] points[j] points[k]] return None # Example usage points = [[0 0] [0 2] [2 0] [2 2] [1 1]] print(no_point_inside_triangle(points))
C# using System; using System.Collections.Generic; class Program { // Function to calculate the area of a triangle given its three vertices static double Area(double x1 double y1 double x2 double y2 double x3 double y3) { return Math.Abs((x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) / 2.0); } // Function to check if a point (x y) is inside a triangle defined by its vertices (x1 y1) (x2 y2) and (x3 y3) static bool IsInsideTriangle(double x1 double y1 double x2 double y2 double x3 double y3 double x double y) { double A = Area(x1 y1 x2 y2 x3 y3); double A1 = Area(x y x2 y2 x3 y3); double A2 = Area(x1 y1 x y x3 y3); double A3 = Area(x1 y1 x2 y2 x y); return Math.Abs(A - (A1 + A2 + A3)) < 1e-9; // Use a small epsilon value for comparison due to floating-point precision } // Function to find three points from a list that do not form a triangle with any other point inside static List<double[]> NoPointInsideTriangle(List<double[]> points) { for (int i = 0; i < points.Count; ++i) { for (int j = i + 1; j < points.Count; ++j) { for (int k = j + 1; k < points.Count; ++k) { bool inside = false; for (int l = 0; l < points.Count; ++l) { if (l != i && l != j && l != k) { if (IsInsideTriangle(points[i][0] points[i][1] points[j][0] points[j][1] points[k][0] points[k][1] points[l][0] points[l][1])) { inside = true; break; } } } if (!inside) { List<double[]> result = new List<double[]> { points[i] points[j] points[k] }; return result; } } } } return new List<double[]>(); // Return an empty list if no such set of points is found } static void Main(string[] args) { List<double[]> points = new List<double[]> { new double[] {0 0} new double[] {0 2} new double[] {2 0} new double[] {2 2} new double[] {1 1} }; List<double[]> result = NoPointInsideTriangle(points); if (result.Count > 0) { Console.WriteLine('Points that do not form a triangle with any other point inside:'); foreach (var point in result) { Console.WriteLine($'({point[0]} {point[1]})'); } } else { Console.WriteLine('No such set of points found.'); } } }
JavaScript // JavaScript equivalent of the Python code above function area(x1 y1 x2 y2 x3 y3) { return Math.abs((x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) / 2.0); } function is_inside_triangle(x1 y1 x2 y2 x3 y3 x y) { const A = area(x1 y1 x2 y2 x3 y3); const A1 = area(x y x2 y2 x3 y3); const A2 = area(x1 y1 x y x3 y3); const A3 = area(x1 y1 x2 y2 x y); return A === A1 + A2 + A3; } function no_point_inside_triangle(points) { for (let i = 0; i < points.length; i++) { for (let j = i + 1; j < points.length; j++) { for (let k = j + 1; k < points.length; k++) { let inside = false; for (let l = 0; l < points.length; l++) { if (l !== i && l !== j && l !== k) { if (is_inside_triangle(points[i][0] points[i][1] points[j][0] points[j][1] points[k][0] points[k][1] points[l][0] points[l][1])) { inside = true; break; } } } if (!inside) { return [points[i] points[j] points[k]]; } } } } return null; } // Example usage const points = [[0 0] [0 2] [2 0] [2 2] [1 1]]; console.log(no_point_inside_triangle(points));
Izvade
[[0 0] [0 2] [1 1]]
Laika sarežģītība: O(n^4), kur n ir punktu skaits. Tas ir tāpēc, ka mums ir jāatkārto visi iespējamie trīsstūri, kas ir n, izvēlieties 3 un pēc tam pārbaudiet, vai katrs no atlikušajiem punktiem atrodas trīsstūrī, kas ir O (n).
Telpas sarežģītība: O(1), jo mēs vienlaikus saglabājam tikai dažus mainīgos.
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