给定N个点可以形成的三角形数量

给定笛卡尔平面上N个点的X和Y坐标。任务是找到可以通过将每个点连接到其他点而形成的具有非零区域的三角形的数量。

例子：

Input: P[] = {{0, 0}, {2, 0}, {1, 1}, {2, 2}}
Output: 3
Possible triangles can be [(0, 0}, (2, 0), (1, 1)], 
[(0, 0), (2, 0), (2, 2)] and [(1, 1), (2, 2), (2, 0)]

Input : P[] = {{0, 0}, {2, 0}, {1, 1}}
Output : 1

笛卡尔坐标系中可能存在的三角形数量中已经讨论了一种朴素的方法

高效的方法：考虑点Z并找到其与其他点的斜率。现在，如果两个点与点Z具有相同的斜率，则意味着这三个点是共线的，它们不能形成三角形。因此，将Z作为其点之一的三角形的数目是从其余点中选择2个点，然后从与Z具有相同斜率的点中减去2个点的方式数目。 N点中的任何点，我们都必须重复一个循环。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// This function returns the required number
// of triangles
int countTriangles(pair P[], int N)
{
    // Hash Map to store the frequency of
    // slope corresponding to a point (X, Y)
    map, int> mp;
    int ans = 0;
  
    // Iterate over all possible points
    for (int i = 0; i < N; i++) {
        mp.clear();
  
        // Calculate slope of all elements
        // with current element
        for (int j = i + 1; j < N; j++) {
            int X = P[i].first - P[j].first;
            int Y = P[i].second - P[j].second;
  
            // find the slope with reduced
            // fraction
            int g = __gcd(X, Y);
            X /= g;
            Y /= g;
            mp[{ X, Y }]++;
        }
        int num = N - (i + 1);
  
        // Total number of ways to form a triangle
        // having one point as current element
        ans += (num * (num - 1)) / 2;
  
        // Subtracting the total number of ways to
        // form a triangle having the same slope or are
        // collinear
        for (auto j : mp)
            ans -= (j.second * (j.second - 1)) / 2;
    }
    return ans;
}
  
// Driver Code to test above function
int main()
{
    pair P[] = { { 0, 0 }, { 2, 0 }, { 1, 1 }, { 2, 2 } };
    int N = sizeof(P) / sizeof(P[0]);
    cout << countTriangles(P, N) << endl;
    return 0;
}

Python3

# Python3 implementation of the above approach 
from collections import defaultdict
from math import gcd
  
# This function returns the 
# required number of triangles 
def countTriangles(P, N): 
  
    # Hash Map to store the frequency of 
    # slope corresponding to a point (X, Y) 
    mp = defaultdict(lambda:0) 
    ans = 0
  
    # Iterate over all possible points 
    for i in range(0, N): 
        mp.clear() 
  
        # Calculate slope of all elements 
        # with current element 
        for j in range(i + 1, N): 
            X = P[i][0] - P[j][0] 
            Y = P[i][1] - P[j][1] 
  
            # find the slope with reduced 
            # fraction 
            g = gcd(X, Y) 
            X //= g 
            Y //= g 
            mp[(X, Y)] += 1
          
        num = N - (i + 1) 
  
        # Total number of ways to form a triangle 
        # having one point as current element 
        ans += (num * (num - 1)) // 2
  
        # Subtracting the total number of 
        # ways to form a triangle having 
        # the same slope or are collinear 
        for j in mp: 
            ans -= (mp[j] * (mp[j] - 1)) // 2
      
    return ans 
  
# Driver Code
if __name__ == "__main__": 
  
    P = [[0, 0], [2, 0], [1, 1], [2, 2]] 
    N = len(P) 
    print(countTriangles(P, N))
      
# This code is contributed by Rituraj Jain

输出：

3

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