直角坐标系中可能三角形的数量

在直角坐标系中给定n个点。计算形成的三角形数量。

例子：

Input  : point[] = {(0, 0), (1, 1), (2, 0), (2, 2)
Output : 3
Three triangles can be formed from above points.

一个简单的解决方案是检查所选择的三个点的行列式是否为非零。以下行列式给出了三角形的面积(也称为克莱默法则)。
角在(x1，y1)，(x2，y2)和(x3，y3)处的三角形的面积为：
${\displaystyle Area = \pm \frac{1}{2}\begin{bmatrix} x1 & y1 & 1\\ x2 & y2 & 1\\ x3 & y3 & 1 \end{bmatrix}}$
我们可以通过采用3点的所有可能组合并找到行列式来解决此问题。

C++

// C++ program to count number of triangles that can
// be formed with given points in 2D
#include 
using namespace std;
 
// A point in 2D
struct Point
{
   int x, y;
};
 
// Returns determinant value of three points in 2D
int det(int x1, int y1, int x2, int y2, int x3, int y3)
{
   return x1*(y2 - y3) - y1*(x2 - x3) + 1*(x2*y3 - y2*x3);
}
 
// Returns count of possible triangles with given array
// of points in 2D.
int countPoints(Point arr[], int n)
{
    int result = 0;  // Initialize result
 
    // Consider all triplets of points given in inputs
    // Increment the result when determinant of a triplet
    // is not 0.
    for (int i=0; i

---

Java

// Java program to count number
// of triangles that can be formed
// with given points in 2D
 
class GFG{
// Returns determinant value
// of three points in 2D
static int det(int x1, int y1, int x2, int y2, int x3, int y3)
{
    return (x1 * (y2 - y3) - y1 *
        (x2 - x3) + 1 * (x2 *
            y3 - y2 * x3));
}
 
// Returns count of possible
// triangles with given array
// of points in 2D.
static int countPoints(int [][]Point, int n)
{
    int result = 0; // Initialize result
 
    // Consider all triplets of
    // points given in inputs
    // Increment the result when
    // determinant of a triplet is not 0.
    for(int i = 0; i < n; i++)
        for(int j = i + 1; j < n; j++)
            for(int k = j + 1; k < n; k++)
                if(det(Point[i][0], Point[i][1],
                    Point[j][0], Point[j][1],
                    Point[k][0], Point[k][1])>=0)
                    result = result + 1;
 
    return result;
}
 
// Driver code
public static void main(String[] args)
{
int Point[][] = {{0, 0},{1, 1},{2, 0},{2, 2}};
int n = Point.length;
System.out.println(countPoints(Point, n));
}
}
// This code is contributed by
// mits

Python

# Python program to count number
# of triangles that can be formed
# with given points in 2D
 
# Returns determinant value
# of three points in 2D
def det(x1, y1, x2, y2, x3, y3):
    return (x1 * (y2 - y3) - y1 *
            (x2 - x3) + 1 * (x2 *
             y3 - y2 * x3))
 
# Returns count of possible
# triangles with given array
# of points in 2D.
def countPoints(Point, n):
 
    result = 0 # Initialize result
 
    # Consider all triplets of
    # points given in inputs
    # Increment the result when
    # determinant of a triplet is not 0.
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if(det(Point[i][0], Point[i][1],
                       Point[j][0], Point[j][1],
                       Point[k][0], Point[k][1])):
                    result = result + 1
 
    return result
 
# Driver code
Point = [[0, 0], [1, 1],
         [2, 0], [2, 2]]
n = len(Point)
print(countPoints(Point, n))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to count number
// of triangles that can be formed
// with given points in 2D
using System;
 
class GFG{
// Returns determinant value
// of three points in 2D
static int det(int x1, int y1, int x2, int y2, int x3, int y3)
{
    return (x1 * (y2 - y3) - y1 *
        (x2 - x3) + 1 * (x2 *
            y3 - y2 * x3));
}
 
// Returns count of possible
// triangles with given array
// of points in 2D.
static int countPoints(int[,] Point, int n)
{
    int result = 0; // Initialize result
 
    // Consider all triplets of
    // points given in inputs
    // Increment the result when
    // determinant of a triplet is not 0.
    for(int i = 0; i < n; i++)
        for(int j = i + 1; j < n; j++)
            for(int k = j + 1; k < n; k++)
                if(det(Point[i,0], Point[i,1], Point[j,0], Point[j,1],Point[k,0], Point[k,1])>=0)
                    result = result + 1;
 
    return result;
}
 
// Driver code
public static void Main()
{
int[,] Point = new int[,] { { 0, 0 }, { 1, 1 }, { 2, 0 }, { 2, 2 } };
int n = Point.Length/Point.Rank;
Console.WriteLine(countPoints(Point, n));
}
}
// This code is contributed by mits

Javascript

 
// JavaScript program to count number
// of triangles that can be formed
// with given points in 2D
 
// Returns determinant value
// of three points in 2D
function det(x1, y1, x2, y2, x3, y3)
{
    return (x1 * (y2 - y3) -
            y1 * (x2 - x3) +
             1 * (x2 * y3 - y2 * x3));
}
   
// Returns count of possible
// triangles with given array
// of points in 2D.
function countPoints(Point, n)
{
     
    // Initialize result
    let result = 0;
   
    // Consider all triplets of
    // points given in inputs
    // Increment the result when
    // determinant of a triplet is not 0.
    for(let i = 0; i < n; i++)
        for(let j = i + 1; j < n; j++)
            for(let k = j + 1; k < n; k++)
                if (det(Point[i][0], Point[i][1],
                    Point[j][0], Point[j][1],
                    Point[k][0], Point[k][1]) >= 0)
                    result = result + 1;
   
    return result;
}
 
// Driver Code
let Point = [ [ 0, 0 ], [ 1, 1 ],
              [ 2, 0 ], [ 2, 2 ] ];
let n = Point.length;
 
document.write(countPoints(Point, n));
 
// This code is contributed by souravghosh0416
 

输出：

3

时间复杂度： $O(n^{3})$ 。