题目传送门：bzoj1013
其实就是设球心在第$ i $维坐标为$ x_i $，第$ i $个点在第$ j $维的坐标为$ a_{i,j} $，然后列一堆方程：
$ \sum_{i=1}^{n} (x_i-a_{1,i})^2 = \sum_{i=1}^{n} (x_i-a_{2,i})^2 = \sum_{i=1}^{n} (x_i-a_{3,i})^2 = ... =\sum_{i=1}^{n} (x_i-a_{n+1,i})^2 $
把完全平方拆开减去同类项就成了n个一次方程，直接上高斯消元（其实就是模拟人手工解方程）。
代码：
#include<cstdio>#include<cmath>#define eps 1e-18double a[20][20],mat[20][20],ans[20];int n;int gauss(){ ???int now=1; ???for(int i=1;i<=n;i++){ ???????int tmp=now; ???????for(;tmp<=n;tmp++) ???????????if(fabs(mat[tmp][i])>eps)break; ???????if(tmp>n)break; ???????if(now<tmp){ ???????????for(int j=1;j<=n+1;j++){ ???????????????double t=mat[now][j]; mat[now][j]=mat[tmp][j]; mat[tmp][j]=t; ???????????} ???????} ???????for(int j=now+1;j<=n;j++){ ???????????double t=mat[j][i]/mat[now][i]; ???????????for(int k=1;k<=n+1;k++) ???????????????mat[j][k]-=t*mat[now][k]; ???????} ???????++now; ???} ???for(int i=now;i<=n;i++) ???????if(fabs(mat[i][n+1])>eps)return 0; ???return 1;}int main(){ ???scanf("%d",&n); ???for(int i=0;i<=n;i++) ???????for(int j=1;j<=n;j++) ???????????scanf("%lf",&a[i][j]); ???for(int i=1;i<=n;i++){ ???????double sum=0; ???????for(int j=1;j<=n;j++){ ???????????sum+=a[0][j]*a[0][j]-a[i][j]*a[i][j]; ???????????mat[i][j]=2*(a[0][j]-a[i][j]); ???????} ???????mat[i][n+1]=sum; ???} ???gauss(); ???for(int i=n;i;i--){ ???????double tmp=mat[i][n+1]; ???????for(int j=1;j<=n;j++) ???????????tmp-=ans[j]*mat[i][j]; ???????ans[i]=tmp/mat[i][i]; ???} ???for(int i=1;i<n;i++) ???????printf("%.3lf ",ans[i]); ???printf("%.3lf\n",ans[n]);}【bzoj1013】[JSOI2008]球形空间产生器sphere
原文地址：https://www.cnblogs.com/quzhizhou/p/10070688.html