令球心坐标为x1,x2...xn,假设当前第i个点坐标为a1,a2...,an，第i+1个点坐标为b1,b2...,bn，则由半径相等可得：
(a1-x1)^2+(a2-x2)^2+...+(an-xn)^2=(b1-x1)^2+(b2-x2)^2+...+(bn-xn)^2
化简可得:
2(a1-b1)x1+2(a2-b2)x2+...+2(an-bn)xn=(a1^2+a2^2+...+an^2-b1^2-b2^2-...-b3^2)
如此可得到n个一元n次方程组，用最简单的高斯消元搞一搞就好了。
By：大奕哥
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int N=15; 4 double a[N][N],x[N][N],sum[N],ans[N]; 5 int n; 6 void Gauss() 7 { 8 ????for(int i=1;i<=n;++i) 9 ????{10 ????????int t=i;11 ????????for(int j=i+1;j<=n;++j)if(a[j][i]>a[t][i])t=j;12 ????????if(t!=i)for(int j=1;j<=n+1;j++)swap(a[i][j],a[t][j]);13 ????????for(int j=i+1;j<=n;++j)14 ????????{15 ????????????double tmp=a[j][i]/a[i][i];16 ????????????for(int k=i;k<=n+1;++k)a[j][k]-=tmp*a[i][k];17 ????????}18 ????}19 ????for(int i=n;i;--i)20 ????{21 ????????ans[i]=a[i][n+1]/a[i][i];22 ????????for(int j=i-1;j;--j)23 ????????{24 ????????????a[j][n+1]-=a[j][i]*ans[i];25 ????????}26 ????}27 }28 int main()29 {30 ????scanf("%d",&n);31 ????for(int i=1;i<=n+1;++i)32 ????{33 ????????for(int j=1;j<=n;++j)34 ????????scanf("%lf",&x[i][j]),sum[i]=sum[i]+x[i][j]*x[i][j];35 ????}36 ????for(int i=1;i<=n;++i)37 ????{38 ????????for(int j=1;j<=n;++j)a[i][j]=2*(x[i][j]-x[i+1][j]);39 ????????a[i][n+1]=sum[i]-sum[i+1];40 ????}41 ????Gauss();42 ????for(int i=1;i<n;++i)printf("%.3lf ",ans[i]);43 ????printf("%.3lf\n",ans[n]);44 ????return 0;45 }BZOJ1013 [JSOI2008]球形空间产生器sphere
原文地址：https://www.cnblogs.com/nbwzyzngyl/p/8269850.html