Donats N punts en l'espai bidimensional, hem de trobar tres punts de manera que el triangle fet escollint aquests punts no hauria de contenir cap altre punt a l'interior. Tots els punts donats no estaran a la mateixa línia, de manera que la solució sempre existirà.
Exemples:
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.
La solució es basa en el fet que si hi ha triangle(s) sense punts dins, podem formar un triangle amb qualsevol punt aleatori entre tots els punts.
Podem resoldre aquest problema buscant els tres punts un per un. El primer punt es pot triar aleatòriament. Després d'escollir el primer punt necessitem dos punts de manera que el seu pendent ha de ser diferent i cap punt ha de situar-se dins del triangle d'aquests tres punts. Ho podem fer escollint el segon punt com a punt més proper al primer i el tercer punt com a segon punt més proper amb la diferent pendent. Per fer-ho, primer iterem sobre tots els punts i triem el punt més proper al primer i el designem com a segon punt del triangle requerit. A continuació, repetim una vegada més per trobar el punt que tingui diferent pendent i la distància més petita que serà el tercer punt del nostre triangle.
Sree RamanujanC++
// C/C++ program to find triangle with no point inside #include
// 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.
# 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
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
<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>
Sortida:
0 0
1 1
0 2
Complexitat temporal: O(n)
Espai auxiliar: O(1)
Aquest article és contribuït per Utkarsh Trivedi .
Enfocament núm. 2: Ús de la força bruta
Aquest codi itera sobre tots els triangles possibles que es poden formar a partir del conjunt de punts donat i comprova si hi ha cap altre punt dins de cada triangle. Si es troba un triangle on no hi ha cap punt dins del codi, retorna el triangle. En cas contrari, retorna Cap.
Algorisme
1. Recorre tots els triangles possibles amb vèrtexs a partir dels punts donats.
2. Per a cada triangle comproveu si algun dels punts restants es troba dins del triangle.
3. Si no hi ha cap punt dins de cap triangle retornen les coordenades del primer triangle trobat.
#include
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.'); } } }
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))
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 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));
Sortida
[[0 0] [0 2] [1 1]]
Complexitat temporal: O(n^4) on n és el nombre de punts. Això es deu al fet que hem d'iterar a través de tots els triangles possibles que és n triar 3 i després comprovar si cadascun dels punts restants es troba dins del triangle que és O (n).
Complexitat espacial: O(1) ja que només emmagatzemem unes quantes variables alhora.
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