P4035 [JSOI2008]球形空间产生器

假设我们设答案坐标 $(x_1,x_2, \cdots ,x_n)$ ，对于两个点 $(a_1,a_2, \cdots ,a_n),(b_1,b_2, \cdots ,b_n)$ 显然有 $\sqrt {(a_1 -x_1)^2 + (a_2-x_2)^2+ \cdots (a_n - x_n)^2} = \sqrt {(b_1 -x_1)^2 + (b_2-x_2)^2+ \cdots (b_n - x_n)^2}$

即 $\sum a_i^2 + \sum 2a_i \times x_i+\sum x_i^2 = \sum b_i^2 + \sum 2b_i \times x_i+\sum x_i^2$

即 $\sum x_i \times 2(a_i-b_i)=\sum a_i^2-\sum b_i^2$ 。

对相邻的两个点列出如此的方程共有 $n$ 个，可知一定有解。

#include <bits/stdc++.h>
#define MAXN 105
using namespace std;
inline int read(){
	int x=0,f=1;
	char ch=getchar();
	while (ch<'0'||ch>'9'){
		if (ch=='-') f=-1;
		ch=getchar();
	}
	while (ch>='0'&&ch<='9'){
		x=(x<<1)+(x<<3)+(ch^'0');
		ch=getchar();
	}
	return x*f;
}
namespace Gauss{
	double c[MAXN][MAXN],ans[MAXN];
	inline void gauss(int n){
		for (register int i=1;i<=n;++i){
			int id=i;
			for (register int j=i+1;j<=n;++j){
				if (fabs(c[j][i])>fabs(c[id][i])) id=j;
			}
			if (id!=i) for (register int j=i;j<=n+1;++j) swap(c[i][j],c[id][j]);
			for (register int j=i+1;j<=n;++j){
				double rate=c[j][i]/c[i][i];
				for (register int k=i;k<=n+1;++k) c[j][k]-=c[i][k]*rate;
			}
		}
		for (register int i=n;i>=1;--i){
			for (register int j=i+1;j<=n;++j){
				c[i][n+1]-=c[j][n+1]*c[i][j];
			}
			c[i][n+1]/=c[i][i];
		}
	}
}
using namespace Gauss;
double a[MAXN][MAXN];
int n;
int main(){
	n=read();
	for (register int i=1;i<=n+1;++i){
		for (register int j=1;j<=n;++j){
			scanf("%lf",&a[i][j]);
		}
	}
	for (register int i=1;i<=n;++i){
		for (register int j=1;j<=n;++j) {
			c[i][j]=2.00*(a[i][j]-a[i+1][j]);
			c[i][n+1]+=a[i][j]*a[i][j]-a[i+1][j]*a[i+1][j];
		}
	}
	gauss(n);
	for (register int i=1;i<=n;++i){
		printf("%.3lf ",c[i][n+1]);
	}
}