「JSOI2008」球形空间产生器sphere - 高斯消元

描述

给定 \(n + 1\) 个 \(n\) 维坐标， 求解 \(n\) 维球心

思路

高斯消元复习题, 将距离表达式中 \(x^2,y^2\) 消掉，得到线性方程组

依次选取第 \(i\) 列进行消元，为了方便，没有消成上三角矩阵然后回带，而是采取了完全消元

#include <bits/stdc++.h>
using namespace std;

double a[20][20], b[20], c[20][20];
int n;

int main() {
  cin >> n;
  for (int i = 1; i <= n + 1; ++ i)
  for (int j = 1; j <= n; ++ j) scanf("%lf", &a[i][j]);
  for (int i = 1; i <= n; ++ i)
    for (int j = 1; j <= n; ++ j) {
      c[i][j] = 2 * (a[i][j] - a[i + 1][j]);
      b[i] += a[i][j] * a[i][j] - a[i + 1][j] * a[i + 1][j];
    }
  for (int i = 1; i <= n; ++ i) {
    for (int j = i; j <= n; ++ j) {
      if (fabs(c[j][i]) > 1e-8) {
        for (int k = 1; k <= n; ++ k) swap(c[i][k], c[j][k]);
        swap(b[i], b[j]);
        break;
      }
    }
    for (int j = 1; j <= n; ++ j) {
      if (i == j) continue;
      double rate = c[j][i] / c[i][i];
      for (int k = i; k <= n; ++ k) c[j][k] -= c[i][k] * rate;
      b[j] -= b[i] * rate;
    }
  }
  for (int i = 1; i < n; ++ i) printf("%.3f ", b[i] / c[i][i]);
  printf("%.3f\n", b[n] / c[n][n]);
}