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

 1 #include <bits/stdc++.h>
 2 #define faststd ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
 3 #define ll long long
 4 using namespace std;
 5 const int MAXN=110;
 6 const double eps=1e-6;
 7 int n;
 8 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
 9 // gauss
10 // O(n^3)
11 int gauss()
12 {
13     int c, r;
14     for (c = 0, r = 0; c < n; c ++ )
15     {
16         int t = r;
17         for (int i = r; i < n; i ++ )
18             if (fabs(a[i][c]) > fabs(a[t][c]))
19                 t = i;
20 
21         if (fabs(a[t][c]) < eps) continue;
22 
23         for (int i = c; i < n + 1; i ++ ) swap(a[t][i], a[r][i]);
24         for (int i = n; i >= c; i -- ) a[r][i] /= a[r][c];
25 
26         for (int i = r + 1; i < n; i ++ )
27             if (fabs(a[i][c]) > eps)
28                 for (int j = n; j >= c; j -- )
29                     a[i][j] -= a[r][j] * a[i][c];
30 
31         r ++ ;
32     }
33 
34     if (r < n)
35     {
36         for (int i = r; i < n; i ++ )
37             if (fabs(a[i][n]) > eps)
38                 return 2;
39         return 1;
40     }
41 
42     for (int i = n - 1; i >= 0; i -- )
43         for (int j = i + 1; j < n; j ++ )
44             a[i][n] -= a[j][n] * a[i][j];
45 
46     return 0;
47 }
48 int main(){
49     faststd
50     cin>>n;
51     for(int i=0;i<=n;i++) {
52         for (int j = 0; j <n; j++)
53             cin >> b[i][j];
54     }
55     for(int i=0;i<n;i++) {
56         double tmp=0;
57         for (int j = 0; j < n; j++) {
58             a[i][j]=2*(b[i][j]-b[i+1][j]);
59             tmp+=b[i][j]*b[i][j]-b[i+1][j]*b[i+1][j];
60         }
61         a[i][n]=tmp;
62     }
63     int t =gauss();
64     for(int i=0;i<n;i++)
65         printf("%.3lf ",a[i][n]);
66     return 0;
67 }

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