bzoj1013 JSOI2008—球形空间产生器sphere

Posted 2020-08-11

www.lydsy.com/JudgeOnline/problem.php?id=1013 (题目链接)

题意：有一个n维的球体，给出球上n+1个点，求出圆心。

Solution 
　　题中给出了对于n维空间点与点之间的距离求法。那么我们将圆心的坐标设为{x1,x2,x3……xn}，那么就可以列出n个n元一次方程。 
　　高斯消元。

代码：

// bzoj1013
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define inf 2147483640
#define eps 1e-7
#define Pi 3.1415926535898
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;

const int maxn=30;
double ans[maxn],f[maxn],a[maxn][maxn];
int n;

double sqr(double x) {
    return x*x;
}
void Gauss() {
    for (int i=1;i<=n;i++) {
        int to=i+1;
        while (to<=n && fabs(a[to][i])<=eps) to++;
        if (to>n) continue;
        for (int j=to;j<=n;j++) {
            double t=a[i][i]/a[j][i];
            for (int k=1;k<=n+1;k++) a[j][k]=a[j][k]*t-a[i][k];
        }
    }
    for (int i=n;i>=1;i--) {
        double t=a[i][n+1];
        for (int j=i+1;j<=n;j++) t-=ans[j]*a[i][j];
        ans[i]=t/a[i][i];
    }
}              
int main() {
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%lf",&f[i]);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++) {
            double t;
            scanf("%lf",&t);
            a[i][j]=2*(t-f[j]);
            a[i][n+1]+=sqr(t)-sqr(f[j]);
        }
    Gauss();
    for (int i=1;i<n;i++) printf("%.3lf ",ans[i]);
    printf("%.3lf",ans[n]);
    return 0;
}