给定点之间的所有对之间的距离的平方和

给定一个由XY平面上N个点的坐标组成的数组arr [] ，任务是找到所有成对点之间的距离的平方和，即(X _i – X _j ) ² +(Y _i – Y _j ) ²对于每个不同的对(i，j) 。

例子：

Input: arr[][] = {{1, 1}, {-1, -1}, {1, -1}, {-1, 1}}
Output: 32
Explanation: 
Distance of 1^st point (1, 1) from the 2^nd, 3^rd and 4^th points are 8, 4 and 4 respectively.
Distance of 2^nd point from the 3^rd and 4^th points are 4 and 4 respectively.
Distance of 3^rd point from the 4^th point is 8.
Therefore, the total distance = (8 + 4 + 4) + (4 + 4) + (8) = 32

Input: arr[][] = {{1, 1}, {1, 1}, {0, 0}}
Output: 4
Explanation:
Distance of 1^st point from the 2^nd and 3^rd points are 0 and 2 respectively.
Distance of 2^nd point from the 3^rd point is 2.
Therefore, the total distance = (0 + 2) + (2) = 4

天真的方法：解决问题的最简单方法是生成给定数组arr [] []的所有可能的不同对，并计算所有点对(X _i ，Y _j )和(X _j ，Y _j )，即(X _i – X _j ) ² +(Y _i – Y _j ) ² ，对于每个不同的对(i，j) 。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：为了优化上述方法，我们的想法是重新组合和并将距离的平方和分成两个和。请按照以下步骤解决问题：

• 初始化变量，例如xq ， yq ， xs和ys 。
• 用零初始化一个变量，例如res ，以存储结果总和。
• 遍历给定的数组，并对每个点{x，y} ，执行以下步骤：
  • 在变量res中添加(i * x ² + i * y ² )的值，这对应于平方距离的相加。
  • 将值(xq – 2 * xs * a)和(yq – 2 * ys * b)相加，以消除(*)在扩展(a – b) ^2时2 * X * Y的影响。
  • 将值a ²和b ²分别添加到变量xq和yq 。
  • 将值a和b分别添加到变量xs和ys中。
  • 将xs和yq值分别添加到变量a ²和b ^2中。
• 完成上述步骤后，打印res的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the sum of squares
// of distance between all distinct pairs
void findSquareSum(
    int Coordinates[][2], int N)
{
    long long xq = 0, yq = 0;
    long long xs = 0, ys = 0;
 
    // Stores final answer
    long long res = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        int a, b;
 
        a = Coordinates[i][0];
        b = Coordinates[i][1];
 
        res += xq;
        res -= 2 * xs * a;
 
        // Adding the effect of this
        // point for all the previous
        // x - points
        res += i * (long long)(a * a);
 
        // Temporarily add the
        // square of x-coordinate
        xq += a * a;
        xs += a;
        res += yq;
        res -= 2 * ys * b;
        res += i * (long long)b * b;
 
        // Add the effect of this point
        // for all the previous y - points
        yq += b * b;
        ys += b;
    }
 
    // Print the desired answer
    cout << res;
}
 
// Driver Code
int main()
{
    int arr[][2] = { { 1, 1 },
                     { -1, -1 },
                     { 1, -1 },
                     { -1, 1 } };
    int N = sizeof(arr) / sizeof(arr[0]);
    findSquareSum(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
 
  // Function to find the sum of squares
  // of distance between all distinct pairs
  static void findSquareSum(
    int Coordinates[][], int N)
  {
    long xq = 0, yq = 0;
    long xs = 0, ys = 0;
 
    // Stores final answer
    long res = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
      int a, b;
 
      a = Coordinates[i][0];
      b = Coordinates[i][1];
 
      res += xq;
      res -= 2 * xs * a;
 
      // Adding the effect of this
      // point for all the previous
      // x - points
      res += i * (long)(a * a);
 
      // Temporarily add the
      // square of x-coordinate
      xq += a * a;
      xs += a;
      res += yq;
      res -= 2 * ys * b;
      res += i * (long)b * b;
 
      // Add the effect of this point
      // for all the previous y - points
      yq += b * b;
      ys += b;
    }
 
    // Print the desired answer
    System.out.println(res);
  }
 
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[][] = { { 1, 1 },
                   { -1, -1 },
                   { 1, -1 },
                   { -1, 1 } };
    int N = arr.length;
    findSquareSum(arr, N);
  }
}
 
// This code is contributed by code_hunt.

Python3

# Python3 program for the above approach
 
# Function to find the sum of squares
# of distance between all distinct pairs
def findSquareSum(Coordinates, N):
    xq , yq = 0, 0
    xs , ys = 0, 0
 
    # Stores final answer
    res = 0
 
    # Traverse the array
    for i in range(N):
 
        a = Coordinates[i][0]
        b = Coordinates[i][1]
 
        res += xq
        res -= 2 * xs * a
 
        # Adding the effect of this
        # point for all the previous
        # x - points
        res += i * (a * a)
 
        # Temporarily add the
        # square of x-coordinate
        xq += a * a
        xs += a
        res += yq
        res -= 2 * ys * b
        res += i * b * b
 
        # Add the effect of this point
        # for all the previous y - points
        yq += b * b
        ys += b
 
    # Print the desired answer
    print (res)
 
# Driver Code
if __name__ == '__main__':
    arr = [ [ 1, 1 ],
         [ -1, -1 ],
         [ 1, -1 ],
         [ -1, 1 ] ]
 
    N = len(arr)
 
    findSquareSum(arr, N)
 
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the sum of squares
// of distance between all distinct pairs
static void findSquareSum(int[,] Coordinates, int N)
{
    long xq = 0, yq = 0;
    long xs = 0, ys = 0;
     
    // Stores final answer
    long res = 0;
     
    // Traverse the array
    for(int i = 0; i < N ; i++)
    {
        int a, b;
         
        a = Coordinates[i, 0];
        b = Coordinates[i, 1];
         
        res += xq;
        res -= 2 * xs * a;
         
        // Adding the effect of this
        // point for all the previous
         
        // x - points
        res += i * (long)(a * a);
         
        // Temporarily add the
        // square of x-coordinate
        xq += a * a;
        xs += a;
        res += yq;
        res -= 2 * ys * b;
        res += i * (long)b * b;
         
        // Add the effect of this point
        // for all the previous y - points
        yq += b * b;
        ys += b;
    }
     
    // Print the desired answer
    Console.Write(res);
}
 
// Driver code
static void Main()
{
    int[,] arr = { { 1, 1 },
                   { -1, -1 },
                   { 1, -1 },
                   { -1, 1 } };
    int N = arr.GetLength(0);
     
    findSquareSum(arr, N);
}
}
 
// This code is contributed by code_hunt

输出：

32

时间复杂度： O(N)
辅助空间： O(1)