二维平面中N个点之间的锤击距离

给定二维平面中的n个点，后跟Xi，Yi描述n个点。任务是计算n点的锤击距离。
注意：锤击距离是每对点之间最短距离的平方之和。

例子：

Input: n = 3
0 1
0 0
1 0
Output: 4

Input: n = 4
1 0
2 0
3 0
4 0
Output: 20

基本方法：由于我们必须找出所有对中最短距离的平方和，因此，我们可以取每个可能的对并计算距离的平方和。

// Pseudo code to find hammered-distance using above approach.
//this will store hammered distance
Distance=0
for(int i=0;i

它的时间复杂度将是O(n ^ 2)。
高效的方法：这个问题可以通过O(N)的时间复杂度来解决。

    $\begin{document} $$Sum=\sum_{i=1}^{n} \sum_{j=1}^{i-1} \left(X_j-X_i \right)^2+\left(Y_j-Y_i \right)^2$$ We can solve separtely for x and y coordinates. For X: $$Sum_x=\sum_{i=1}^{n} \sum_{j=1}^{i-1} \left(X_j-X_i \right)^2$$ $$Sum_x= \sum_{i=1}^{n} \sum_{j=1}^{i-1} \left(X_j^2+X_i^2 -2 \cdot X_i \cdot X_j \right)$$ Now expand the summation part, We can write this equation as- $$Sum_x=\sum_{i=1}^{n} \left( (i-1)*X_i^2+\sum_{j=1}^{i-1}X_j^2-2 \cdot X_i\cdot \sum_{j=1}^{i-1} X_j \right)$$ $\sum_{j=1}^{i-1}X_j$ This is commulative sum of square of points upto i-1.So, this can be calculated in linear time. Similarly, This can also be calculated in linear time. $2 \cdot X_i\cdot \sum_{j=1}^{i-1} X_j$ \end{document}$

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
#define ll long long int
using namespace std;
 
// Function calculate cummalative sum
// of x, y, x^2, y^2 coordinates.
void cumm(vector& x, vector& y,
        vector& cummx, vector& cummy,
        vector& cummx2, vector& cummy2, ll n)
{
    for (int i = 1; i <= n; i++) {
        cummx[i] = cummx[i - 1] + x[i];
        cummy[i] = cummy[i - 1] + y[i];
        cummx2[i] = cummx2[i - 1] + x[i] * x[i];
        cummy2[i] = cummy2[i - 1] + y[i] * y[i];
    }
}
 
// Function ot calculate the hammered distance
int calHammeredDistance(int n, vector& x, vector& y)
{
    // cummx conatins cummulative sum of x
    // cummy conatins cummulative sum of y
    vector cummx(n + 1, 0), cummy(n + 1, 0);
 
    // cummx2 conatins cummulative sum of x^2
    // cummy2 conatins cummulative sum of y^2
    vector cummx2(n + 1, 0), cummy2(n + 1, 0);
 
    // calculate cummalative of x
    //, y, x^2, y^2, because these terms
    // required in formula to reduce complexity.
 
    // this function calculate all required terms.
    cumm(x, y, cummx, cummy, cummx2, cummy2, n);
 
    // hdx calculate hammer distance for x coordinate
    // hdy calculate hammer distance for y coordinate
    ll hdx = 0, hdy = 0;
 
    for (int i = 1; i <= n; i++) {
 
        // came from formula describe in explanation
        hdx += (i - 1) * x[i] * x[i] + cummx2[i - 1]
            - 2 * x[i] * cummx[i - 1];
 
        // came from formula describe in explanation
        hdy += (i - 1) * y[i] * y[i] + cummy2[i - 1]
            - 2 * y[i] * cummy[i - 1];
    }
 
    // total is the sum of both x and y.
    ll total = hdx + hdy;
    return total;
}
 
// Driver code
int main()
{
    // number of points
    int n = 3;
 
    // x contains the x coordinates
    // y conatins the y coordinates
    //and converting the size to n+1
    vector x = {0, 0, 1, 0};
    vector y = {1, 0, 0, 0};
 
    cout << calHammeredDistance(n, x, y);
 
    return 0;
}

Java

// Java implementation of above approach
 
 
class GFG{
  
// Function calculate cummalative sum
// of x, y, x^2, y^2 coordinates.
static void cumm(int [] x, int [] y,
        int [] cummx, int [] cummy,
        int [] cummx2, int [] cummy2, int n)
{
    for (int i = 1; i <= n; i++) {
        cummx[i] = cummx[i - 1] + x[i];
        cummy[i] = cummy[i - 1] + y[i];
        cummx2[i] = cummx2[i - 1] + x[i] * x[i];
        cummy2[i] = cummy2[i - 1] + y[i] * y[i];
    }
}
  
// Function ot calculate the hammered distance
static int calHammeredDistance(int n, int [] x, int [] y)
{
    // cummx conatins cummulative sum of x
    // cummy conatins cummulative sum of y
    int []cummx = new int[n + 1];
    int []cummy = new int[n + 1];
  
    // cummx2 conatins cummulative sum of x^2
    // cummy2 conatins cummulative sum of y^2
    int []cummx2 = new int[n + 1];
    int []cummy2 = new int[n + 1];
  
    // calculate cummalative of x
    //, y, x^2, y^2, because these terms
    // required in formula to reduce complexity.
  
    // this function calculate all required terms.
    cumm(x, y, cummx, cummy, cummx2, cummy2, n);
  
    // hdx calculate hammer distance for x coordinate
    // hdy calculate hammer distance for y coordinate
    int hdx = 0, hdy = 0;
  
    for (int i = 1; i <= n; i++) {
  
        // came from formula describe in explanation
        hdx += (i - 1) * x[i] * x[i] + cummx2[i - 1]
               - 2 * x[i] * cummx[i - 1];
  
        // came from formula describe in explanation
        hdy += (i - 1) * y[i] * y[i] + cummy2[i - 1]
               - 2 * y[i] * cummy[i - 1];
    }
  
    // total is the sum of both x and y.
    int total = hdx + hdy;
    return total;
}
  
// Driver code
public static void main(String[] args)
{
    // number of points
    int n = 3;
  
    // x contains the x coordinates
    // y conatins the y coordinates
    int []x = new int[n + 1];
    int []y = new int[n + 1];
    x[2] = 1;
    y[0] = 1;
  
    System.out.print(calHammeredDistance(n, x, y));
  
}
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 implementation of the
# above approach
 
# Function calculate cummalative sum
# of x, y, x^2, y^2 coordinates.
def cumm(x, y, cummx, cummy,
               cummx2, cummy2, n):
 
    for i in range(1, n+1):
        cummx[i] = cummx[i - 1] + x[i]
        cummy[i] = cummy[i - 1] + y[i]
        cummx2[i] = cummx2[i - 1] + x[i] * x[i]
        cummy2[i] = cummy2[i - 1] + y[i] * y[i]
 
# Function ot calculate the
# hammered distance
def calHammeredDistance(n, x, y):
 
    # cummx conatins cummulative sum of x
    # cummy conatins cummulative sum of y
    cummx = [0] * (n + 1)
    cummy = [0] * (n + 1)
 
    # cummx2 conatins cummulative sum of x^2
    # cummy2 conatins cummulative sum of y^2
    cummx2 = [0] * (n + 1)
    cummy2 = [0] * (n + 1)
 
    # calculate cumulative of x , y, x^2, y^2,
    # because these terms are required in the
    # formula to reduce complexity.
 
    # This function calculate all required terms.
    cumm(x, y, cummx, cummy, cummx2, cummy2, n)
 
    # hdx calculate hammer distance for x coordinate
    # hdy calculate hammer distance for y coordinate
    hdx, hdy = 0, 0
 
    for i in range(1, n + 1):
 
        # came from formula describe in explanation
        hdx += ((i - 1) * x[i] * x[i] + cummx2[i - 1] -
                             2 * x[i] * cummx[i - 1])
 
        # came from formula describe in explanation
        hdy += ((i - 1) * y[i] * y[i] + cummy2[i - 1] -
                             2 * y[i] * cummy[i - 1])
     
    # total is the sum of both x and y.
    total = hdx + hdy
    return total
 
# Driver Code
if __name__ == "__main__":
 
    # number of points
    n = 3
 
    # x contains the x coordinates
    # y conatins the y coordinates
    x = [0, 0, 1, 0]
    y = [1, 0, 0, 0]
 
    print(calHammeredDistance(n, x, y))
 
# This code is contributed by Rituraj Jain

C#

// C# implementation of above approach
using System;
 
class GFG{
   
// Function calculate cummalative sum
// of x, y, x^2, y^2 coordinates.
static void cumm(int [] x, int [] y,
        int [] cummx, int [] cummy,
        int [] cummx2, int [] cummy2, int n)
{
    for (int i = 1; i <= n; i++) {
        cummx[i] = cummx[i - 1] + x[i];
        cummy[i] = cummy[i - 1] + y[i];
        cummx2[i] = cummx2[i - 1] + x[i] * x[i];
        cummy2[i] = cummy2[i - 1] + y[i] * y[i];
    }
}
   
// Function ot calculate the hammered distance
static int calHammeredDistance(int n, int [] x, int [] y)
{
    // cummx conatins cummulative sum of x
    // cummy conatins cummulative sum of y
    int []cummx = new int[n + 1];
    int []cummy = new int[n + 1];
   
    // cummx2 conatins cummulative sum of x^2
    // cummy2 conatins cummulative sum of y^2
    int []cummx2 = new int[n + 1];
    int []cummy2 = new int[n + 1];
   
    // calculate cummalative of x
    //, y, x^2, y^2, because these terms
    // required in formula to reduce complexity.
   
    // this function calculate all required terms.
    cumm(x, y, cummx, cummy, cummx2, cummy2, n);
   
    // hdx calculate hammer distance for x coordinate
    // hdy calculate hammer distance for y coordinate
    int hdx = 0, hdy = 0;
   
    for (int i = 1; i <= n; i++) {
   
        // came from formula describe in explanation
        hdx += (i - 1) * x[i] * x[i] + cummx2[i - 1]
               - 2 * x[i] * cummx[i - 1];
   
        // came from formula describe in explanation
        hdy += (i - 1) * y[i] * y[i] + cummy2[i - 1]
               - 2 * y[i] * cummy[i - 1];
    }
   
    // total is the sum of both x and y.
    int total = hdx + hdy;
    return total;
}
   
// Driver code
public static void Main(String[] args)
{
    // number of points
    int n = 3;
   
    // x contains the x coordinates
    // y conatins the y coordinates
    int []x = new int[n + 1];
    int []y = new int[n + 1];
    x[2] = 1;
    y[0] = 1;
   
    Console.Write(calHammeredDistance(n, x, y)); 
}
}
 
// This code is contributed by PrinciRaj1992

输出

2