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。
A triplet of points (P1, P2, P3) is said to be equidistant when the distance between P1 and P2 is the same as of the distance between P1 and P3.
注意：点的顺序很重要，即(P1，P2，P3)与(P2，P3，P1)不同。
例子：
Input: arr = [[0, 0], [1, 0], [2, 0]]
Output: 2
Explanation:
Since the order of the points matters, we have two different sets of points [[1, 0], [0, 0], [2, 0]] and [[1, 0], [2, 0], [0, 0]] in which the points are equidistant.
Input: arr = [[1, 1], [1, 3], [2, 0]]
Output: 0
Explanation:
It is not possible to get any such triplet in which the points are equidistant.
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 29C#
// 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 输出：
2时间复杂度： O(N 2 )
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