球形空间产生器

球形空间产生器

# 题意

n维度空间的球形，知道球面上的n+1个点的坐标，求球心

# 题解

设半径为r，球心为(x₁ , x₂ , ...... , x_n)

那么会满足

(a₁₁ - x₁ )² + (a₁₂ -x₂)² + ...... + (a_1n - x_n)² = r² 

(a₂₁ - x₁ )² + (a₂₂ -x₂)² + ...... + (a_2n - x_n)² = r² 

......

(a_n1 - x₁ )² + (a_n2 -x₂)² + ...... + (a_nn - x_n)² = r² 

(a_{n+1 1} - x₁ )² + (a_{n+1 2} -x₂)² + ...... + (a_{n+1 n} - x_n)² = r² 

有n+1个点的坐标，某两个能够消去一个x_i的二次项,同时右边的r²消成0，进行n次即可

单独看某两个的x₁ 项相减 ,(a_i1 - x₁)² - (a_{i+1 1} - x₁)² = 0

展开化简得到 (a_i1² - a_{i+1 1}²) + 2 · x₁ · (a_{i+1 1} -a_i1) =0

即(a_i1² - a_{i+1 1}²) = 2 · (a_i1 - a_{i+1 1} ) · x₁ 

可以由此得到每两个方程合并后的方程为 ∑(j=1:n) 2 · (a_ij - a_i+1,j) · x_j = (a_ij² - a_{i+1 j}²)                   

然后高斯消元得圆心坐标即可

#include <bits/stdc++.h>
#define faststd ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll long long
using namespace std;
const int MAXN=110;
const double eps=1e-6;
int n;
double a[MAXN][MAXN],b[MAXN][MAXN];// 最后一列开始存的是方程右边常数，运算后解
// gauss
// O(n^3)
int gauss()
{
    int c, r;
    for (c = 0, r = 0; c < n; c ++ )
    {
        int t = r;
        for (int i = r; i < n; i ++ )
            if (fabs(a[i][c]) > fabs(a[t][c]))
                t = i;

        if (fabs(a[t][c]) < eps) continue;

        for (int i = c; i < n + 1; i ++ ) swap(a[t][i], a[r][i]);
        for (int i = n; i >= c; i -- ) a[r][i] /= a[r][c];

        for (int i = r + 1; i < n; i ++ )
            if (fabs(a[i][c]) > eps)
                for (int j = n; j >= c; j -- )
                    a[i][j] -= a[r][j] * a[i][c];

        r ++ ;
    }

    if (r < n)
    {
        for (int i = r; i < n; i ++ )
            if (fabs(a[i][n]) > eps)
                return 2;
        return 1;
    }

    for (int i = n - 1; i >= 0; i -- )
        for (int j = i + 1; j < n; j ++ )
            a[i][n] -= a[j][n] * a[i][j];

    return 0;
}
int main(){
    faststd
    cin>>n;
    for(int i=0;i<=n;i++) {
        for (int j = 0; j <n; j++)
            cin >> b[i][j];
    }
    for(int i=0;i<n;i++) {
        double tmp=0;
        for (int j = 0; j < n; j++) {
            a[i][j]=2*(b[i][j]-b[i+1][j]);
            tmp+=b[i][j]*b[i][j]-b[i+1][j]*b[i+1][j];
        }
        a[i][n]=tmp;
    }
    int t =gauss();
    for(int i=0;i<n;i++)
        printf("%.3lf ",a[i][n]);
    return 0;
}