给定二维空间中的N个点，我们需要找到三个点，以使通过选择这些点而形成的三角形内部不应包含任何其他点。所有给定的点将不在同一条线上，因此解决方案将始终存在。
例子：
在上图中，可以通过选择这些三元组[(0，0)，(2，0)，(1，1)] [[0，0)，(1，1)，( 0，2)] [((1，1)，(2，0)，(2，2)] [(1，1)，(0，2)，(2，2)]因此，上述任何一个三元组都可以成为最终答案。
该解决方案基于以下事实：如果存在内部不带任何点的三角形，则我们可以形成一个在所有点之间具有任意随机点的三角形。
我们可以通过一一搜索所有三个点来解决此问题。第一点可以随机选择。选择第一个点后，我们需要两个点，以使它们的斜率应该不同，并且没有点应位于这三个点的三角形内。我们可以通过选择第二个点作为与第一个点的最近点，以及选择第三个点作为具有不同斜率的第二个最近点来做到这一点。为此，我们首先遍历所有点，然后选择最接近第一个点的点，并将其指定为所需三角形的第二个点。然后，我们再迭代一次，以找到斜率不同且距离最小的点，这将是三角形的第三个点。
C++
// 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 YuvarajC#
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输出：
0, 0
1, 1
0, 2