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📜  点 B 和 C 与 A 等距的三元组 (A, B, C) 的总数

📅  最后修改于: 2021-10-23 08:06:32             🧑  作者: Mango

给定一个包含N个点的数组arr ,任务是找到其中点等距的三元组的总数。

注意:点的顺序很重要,即(P1,P2,P3)与(P2,P3,P1)不同。

例子:

方法:为了解决上面提到的问题,我们知道三元组的顺序很重要,因此满足等距点对条件的相同三元组可能有不止一个排列

  • 首先,我们将计算具有等距点的三元组的所有排列。
  • 对列表中每个不同的三元组点重复相同的过程。为了计算距离,我们将使用各个坐标之间距离的平方。
  • 使用hashmap来存储单个三元组的不同数量的等距点对。
  • 一旦我们计算出对的总数,我们就会计算所需的排列。我们对所有不同的三元组重复此过程,并将所有排列添加到我们的结果中。

下面是上述方法的实现:

C++
// C++ implementation to Find
// the total number of Triplets
// in which the points are Equidistant
 
#include 
using namespace std;
 
// function to count such triplets
int numTrip(vector >& points)
{
    int res = 0;
 
    // Iterate over all the points
    for (int i = 0; i < points.size(); ++i) {
 
        unordered_map
            map(points.size());
 
        // Iterate over all points other
        // than the current point
        for (int j = 0; j < points.size(); ++j) {
 
            if (j == i)
                continue;
 
            int dy = points[i].second
                     - points[j].second;
            int dx = points[i].first
                     - points[j].first;
 
            // Compute squared euclidean distance
            // for the current point
            int key = dy * dy;
            key += dx * dx;
 
            map[key]++;
        }
 
        for (auto& p : map)
 
            // Compute nP2 that is n * (n - 1)
            res += p.second * (p.second - 1);
    }
 
    // Return the final result
    return res;
}
 
// Driver code
int main()
{
    vector > mat
        = { { 0, 0 }, { 1, 0 }, { 2, 0 } };
 
    cout << numTrip(mat);
 
    return 0;
}


Java
// Java implementation to find the total
// number of Triplets in which the
// points are Equidistant
import java.util.*;
 
@SuppressWarnings("unchecked")
class GFG{
  
static class pair
{
    public int first, second;
      
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
   
// Function to size() such triplets
static int numTrip(ArrayList points)
{
    int res = 0;
     
    // Iterate over all the points
    for(int i = 0; i < points.size(); ++i)
    {
         
        HashMap map = new HashMap<>();
        
        // Iterate over all points other
        // than the current point
        for(int j = 0; j < points.size(); ++j)
        {
            if (j == i)
                continue;
                  
            int dy = ((pair)points.get(i)).second -
                     ((pair)points.get(j)).second;
            int dx = ((pair)points.get(i)).first -
                     ((pair)points.get(j)).first;
   
            // Compute squared euclidean distance
            // for the current point
            long key = dy * dy;
            key += dx * dx;
              
            if (map.containsKey(key))
            {
                map.put(key, map.get(key) + 1);
            }
            else
            {
                map.put(key, 1);
            }
        }
          
        for(int p : map.values())
         
            // Compute nP2 that is n * (n - 1)
            res += p * (p - 1);
    }
     
    // Return the final result
    return res;
}
   
// Driver code
public static void main(String []args)
{
    ArrayList mat = new ArrayList();
    mat.add(new pair(0, 0));
    mat.add(new pair(1, 0));
    mat.add(new pair(2, 0));
     
    System.out.println(numTrip(mat));
}
}
 
// This code is contributed by rutvik_56


Python3
# Python3 implementation to find
# the total number of Triplets
# in which the points are Equidistant
  
# Function to count such triplets
def numTrip(points):
     
    res = 0
     
    # Iterate over all the points
    for i in range(len(points)):
        map = {}
         
        # Iterate over all points other
        # than the current point
        for j in range(len(points)):
            if (j == i):
                continue
             
            dy = points[i][1] - points[j][1]
            dx = points[i][0] - points[j][0]
  
            # Compute squared euclidean distance
            # for the current point
            key = dy * dy
            key += dx * dx
  
            map[key] = map.get(key, 0) + 1
  
        for p in map:
             
            # Compute nP2 that is n * (n - 1)
            res += map[p] * (map[p] - 1)
  
    # Return the final result
    return res
  
# Driver code
if __name__ == '__main__':
     
    mat = [ [ 0, 0 ],
            [ 1, 0 ],
            [ 2, 0 ] ]
  
    print (numTrip(mat))
  
# This code is contributed by mohit kumar 29


C#
// C# implementation to find the total
// number of Triplets in which the
// points are Equidistant
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
 
class pair
{
    public int first, second;
     
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
  
// Function to count such triplets
static int numTrip(ArrayList points)
{
    int res = 0;
     
    // Iterate over all the points
    for(int i = 0; i < points.Count; ++i)
    {
         
        Dictionary map = new Dictionary();
       
        // Iterate over all points other
        // than the current point
        for(int j = 0; j < points.Count; ++j)
        {
            if (j == i)
                continue;
                 
            int dy = ((pair)points[i]).second -
                     ((pair)points[j]).second;
            int dx = ((pair)points[i]).first -
                     ((pair)points[j]).first;
  
            // Compute squared euclidean distance
            // for the current point
            int key = dy * dy;
            key += dx * dx;
             
            if (map.ContainsKey(key))
            {
                map[key]++;
            }
            else
            {
                map[key] = 1;
            }
        }
         
        foreach(int p in map.Values)
         
            // Compute nP2 that is n * (n - 1)
            res += p * (p - 1);
    }
  
    // Return the final result
    return res;
}
  
// Driver code
public static void Main(string []args)
{
    ArrayList mat = new ArrayList(){ new pair(0, 0),
                                     new pair(1, 0),
                                     new pair(2, 0) };
   
    Console.Write(numTrip(mat));
}
}
 
// This code is contributed by pratham76


Javascript


输出:
2

时间复杂度: O(N 2 )

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