给定矩阵mat [] []，该矩阵mat由[ x，y}形式的N对组成，每对表示N个点的坐标，任务是找到到所有点的欧几里得距离的最小和。
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，因此我们的想法是计算所有点的中值。几何中位数将中位数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，计算所有给定坐标的质心。
- 求出质心到所有点的欧几里得距离。
- 计算这些距离的总和并作为答案。
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 KatiyarPython3
# 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 29C#
// 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输出：
4
时间复杂度： O(N)
辅助空间： O(1)