到所有给定点的最小欧氏距离总和

给定一个矩阵mat[][]由N对形式{x, y} 组成，每对表示N个点的坐标，任务是找到到所有点的欧几里德距离的最小总和。

例子：

Input: mat[][] = { { 0, 1}, { 1, 0 }, { 1, 2 }, { 2, 1 }} 
Output: 4 
Explanation: 
Average of the set of points, i.e. Centroid = ((0+1+1+2)/4, (1+0+2+1)/4) = (1, 1). 
Euclidean distance of each point from the centroid are {1, 1, 1, 1} 
Sum of all distances = 1 + 1 + 1 + 1 = 4
Input: mat[][] = { { 1, 1}, { 3, 3 }} 
Output: 2.82843

方法：
由于任务是最小化到所有点的欧几里德距离，所以想法是计算所有点的中值。几何中值将中值的概念推广到更高维度

请按照以下步骤解决问题：

• 通过获取点的平均值来计算所有给定坐标的质心。
• 找到所有点与质心的欧几里德距离。
• 计算这些距离的总和并打印为答案。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to calculate Euclidean distance
double find(double x, double y,
            vector >& p)
{
 
    double mind = 0;
    for (int i = 0; i < p.size(); i++) {
 
        double a = p[i][0], b = p[i][1];
        mind += sqrt((x - a) * (x - a)
                     + (y - b) * (y - b));
    }
 
    return mind;
}
 
// Function to calculate the minimum sum
// of the euclidean distances to all points
double getMinDistSum(vector >& p)
{
 
    // Calculate the centroid
    double x = 0, y = 0;
    for (int i = 0; i < p.size(); i++) {
        x += p[i][0];
        y += p[i][1];
    }
    x = x / p.size();
    y = y / p.size();
 
    // Calculate distance of all
    // points
    double mind = find(x, y, p);
 
    return mind;
}
 
// Driver Code
int main()
{
 
    // Initializing the points
    vector > vec
        = { { 0, 1 }, { 1, 0 }, { 1, 2 }, { 2, 1 } };
 
    double d = getMinDistSum(vec);
    cout << d << endl;
 
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{
 
// Function to calculate Euclidean distance
static double find(double x, double y,
                   int [][] p)
{
    double mind = 0;
     
    for(int i = 0; i < p.length; i++)
    {
        double a = p[i][0], b = p[i][1];
        mind += Math.sqrt((x - a) * (x - a) +
                          (y - b) * (y - b));
    }
    return mind;
}
 
// Function to calculate the minimum sum
// of the euclidean distances to all points
static double getMinDistSum(int [][]p)
{
     
    // Calculate the centroid
    double x = 0, y = 0;
    for(int i = 0; i < p.length; i++)
    {
        x += p[i][0];
        y += p[i][1];
    }
     
    x = x / p.length;
    y = y / p.length;
 
    // Calculate distance of all
    // points
    double mind = find(x, y, p);
 
    return mind;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Initializing the points
    int [][]vec = { { 0, 1 }, { 1, 0 },
                    { 1, 2 }, { 2, 1 } };
 
    double d = getMinDistSum(vec);
     
    System.out.print(d + "\n");
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach
from math import sqrt
 
# Function to calculate Euclidean distance
def find(x, y, p):
 
    mind = 0
    for i in range(len(p)):
        a = p[i][0]
        b = p[i][1]
        mind += sqrt((x - a) * (x - a) +
                     (y - b) * (y - b))
                      
    return mind
 
# Function to calculate the minimum sum
# of the euclidean distances to all points
def getMinDistSum(p):
 
    # Calculate the centroid
    x = 0
    y = 0
     
    for i in range(len(p)):
        x += p[i][0]
        y += p[i][1]
         
    x = x // len(p)
    y = y // len(p)
 
    # Calculate distance of all
    # points
    mind = find(x, y, p)
 
    return mind
 
# Driver Code
if __name__ == '__main__':
 
    # Initializing the points
    vec = [ [ 0, 1 ], [ 1, 0 ],
            [ 1, 2 ], [ 2, 1 ] ]
 
    d = getMinDistSum(vec)
    print(int(d))
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to calculate Euclidean distance
static double find(double x, double y,
                   int [,] p)
{
    double mind = 0;
     
    for(int i = 0; i < p.GetLength(0); i++)
    {
        double a = p[i,0], b = p[i,1];
        mind += Math.Sqrt((x - a) * (x - a) +
                          (y - b) * (y - b));
    }
    return mind;
}
 
// Function to calculate the minimum sum
// of the euclidean distances to all points
static double getMinDistSum(int [,]p)
{
     
    // Calculate the centroid
    double x = 0, y = 0;
    for(int i = 0; i < p.GetLength(0); i++)
    {
        x += p[i,0];
        y += p[i,1];
    }
     
    x = x / p.Length;
    y = y / p.Length;
 
    // Calculate distance of all
    // points
    double mind = find(x, y, p);
 
    return mind;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Initializing the points
    int [,]vec = { { 0, 1 }, { 1, 0 },
                    { 1, 2 }, { 2, 1 } };
 
    int d = (int)getMinDistSum(vec);
     
    Console.Write(d + "\n");
}
}
 
// This code is contributed by Rohit_ranjan

Javascript

 
// JavaScript program for the above approach
 
// Function to calculate Euclidean distance
function find(x, y, p)
{
    let mind = 0;
       
    for(let i = 0; i < p.length; i++)
    {
        let a = p[i][0], b = p[i][1];
        mind += Math.sqrt((x - a) * (x - a) +
                          (y - b) * (y - b));
    }
    return mind;
}
   
// Function to calculate the minimum sum
// of the euclidean distances to all polets
function getMinDistSum(p)
{
       
    // Calculate the centroid
    let x = 0, y = 0;
    for(let i = 0; i < p.length; i++)
    {
        x += p[i][0];
        y += p[i][1];
    }
       
    x = x / p.length;
    y = y / p.length;
   
    // Calculate distance of all
    // polets
    let mind = find(x, y, p);
   
    return mind;
}
     
// Driver Code
         
    // Initializing the points
    let vec = [[ 0, 1 ], [ 1, 0 ],
               [ 1, 2 ], [ 2, 1 ]];
   
    let d = getMinDistSum(vec);
       
    document.write(d);
                      

输出：

4

时间复杂度： O(N)
辅助空间： O(1)