Adott N pont a 2-dimenziós térben, három olyan pontot kell találnunk, hogy a pontok kiválasztásával létrehozott háromszög ne tartalmazzon más pontot. Nem minden adott pont ugyanazon az egyenesen lesz, így a megoldás mindig létezik.
Példák:
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.
A megoldás azon alapul, hogy ha létezik olyan háromszög(ek), amelyeken belül nincs pont, akkor az összes pont közül tetszőleges ponttal alkothatunk háromszöget.
Ezt a problémát úgy tudjuk megoldani, hogy mindhárom pontot egyenként keressük. Az első pont véletlenszerűen kiválasztható. Az első pont kiválasztása után két olyan pontra van szükségünk, amelyek meredeksége eltérő legyen, és egyetlen pont sem helyezkedhet el a három pont háromszögén belül. Ezt úgy tehetjük meg, hogy a második pontot választjuk az elsőhöz legközelebbi pontnak, a harmadikat pedig a második legközelebbi pontnak különböző meredekséggel. Ehhez először iterálunk minden pontot, és kiválasztjuk az elsőhöz legközelebb eső pontot, és ezt jelöljük ki a kívánt háromszög második pontjaként. Ezután ismételjük meg még egyszer, hogy megtaláljuk azt a pontot, amelynek különböző meredeksége van, és amelynek a legkisebb távolsága van, amely háromszögünk harmadik pontja lesz.
hány hónap a q3C++
// 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>
Kimenet:
0 0
1 1
0 2
Időbeli összetettség: On)
Segédtér: O(1)
Ennek a cikknek a közreműködője Utkarsh Trivedi .
2. megközelítés: Brute Force használata
Ez a kód az adott ponthalmazból kialakítható összes lehetséges háromszöget iterálja, és ellenőrzi, hogy az egyes háromszögeken belül van-e más pont. Ha olyan háromszöget találunk, ahol nincs pont a kódon belül, akkor a háromszöget adja vissza. Ellenkező esetben a None-t adja vissza.
Algoritmus
1. Ismételje meg az összes lehetséges háromszöget, amelynek csúcsai a megadott pontokból származnak.
2. Minden háromszögnél ellenőrizze, hogy a fennmaradó pontok valamelyike a háromszög belsejében található-e.
3. Ha egyetlen háromszögben sem található pont, adja vissza az első talált háromszög koordinátáit.
#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));
Kimenet
[[0 0] [0 2] [1 1]]
Időbonyolultság: O(n^4), ahol n a pontok száma. Ennek az az oka, hogy végig kell iterálnunk az összes lehetséges háromszöget, amely n, válassza a 3-at, majd ellenőrizze, hogy a fennmaradó pontok mindegyike az O(n) háromszögön belül van-e.
Térbonyolultság: O(1), mivel egyszerre csak néhány változót tárolunk.
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