球形空间产生器

luogu 4035

金牌导航 高斯消元-1

题目大意

给出n+1个n维的点，让你求一个点，使该点到所有点欧几里得距离相等

输入样例

2
0.0 0.0
-1.0 1.0
1.0 0.0

输出样例

0.500 1.500

数据范围

$1⩽N⩽10$

解题思路

因为所有点到该点的欧几里得距离相等，那么可以设以下方程（同时去根号）
$$(a_{i,1}-x_1{)}^2+(a_{i,2}-x_2{)}^2+...+(a_{i,n}-x_n{)}^2=(a_{i-1,1}-x_1{)}^2+(a_{i-1,2}-x_2{)}^2+...+(a_{i-1,n}-x_n{)}^2$$
解得
$$2x_1(a_{i,1}-a_{i-1,1})+2x_2(a_{i,2}-a_{i-1,2})+...+2x_n(a_{i,n}-a_{i-1,n})=a_{i,1}^2-a_{i-1,1}^2+a_{i,2}^2-a_{i-1,2}^2+...+a_{i,n}^2-a_{i-1,n}^2$$
同理，推出另外n-1个式子，然后高斯消元

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define abss(x) (x < 0? -x: x)
using namespace std;
int n;
double a[20][20], s[20][20];
void solve()//高斯消元
{for (int i = 1; i <= n; ++i){int h = i;for (int j = i + 1; j <= n; ++j)if (abss(s[j][i]) > abss(s[h][i])) h = j;if (abss(s[h][i]) - 0.0 < 1e-6) continue;if (h != i)for (int j = i; j <= n + 1; ++j)swap(s[i][j], s[h][j]);double g = s[i][i];for (int j = i; j <= n + 1; ++j)s[i][j] /= g;for (int j = 1; j <= n; ++j)if (j != i){g = s[j][i];for (int k = i; k <= n + 1; ++k)s[j][k] -= s[i][k] * g;}}return;
}
int main()
{scanf("%d", &n);for (int i = 1; i <= n + 1; ++i)for (int j = 1; j <= n; ++j){scanf("%lf", &a[i][j]);s[i - 1][j] = 2 * (a[i][j] - a[i - 1][j]);s[i - 1][n + 1] += a[i][j] * a[i][j] - a[i - 1][j] * a[i - 1][j];//建方程}solve();for (int i = 1; i <= n; ++i)printf("%.3lf ", s[i][n + 1]);return 0;
}