【BZOJ】1013: [JSOI2008]球形空间产生器sphere
【BZOJ】1013: [JSOI2008]球形空间产生器sphere
题意:给n+1个n维的点的坐标,要你求出一个到这n+1个点距离相等的点的坐标;
思路:高斯消元即第i个点和第i+1个点处理出一个式子,这样n+1个点正好有n个系数的n元变量,即可求解。
式子:Σ( (a[i][j] - x[j])^2 ) = Σ( a[i+1][j] - x[j])^2 )
=> Σ( x[j]*[2*(a[i+1][j]-a[i][j])] ) = Σ(a[i+1][j]*a[i+1][j] - a[i][j]*a[i][j]);直接预处理即可;
注意:在Gauss处理出上三角阵的过程中,每次要选出主对角线绝对值最大的行作为参考行,貌似是精度问题。还有就是归零的过程中,要变成参考行再消,为了不出现除0的情况。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<cstdio> #include<cstring> #include<string.h> #include<algorithm> #include<map> #include<queue> #include<vector> #include<cmath> #include<stdlib.h> #include<time.h> using namespace std; typedef long long ll; #define rep0(i,l,r) for(int i = (l);i < (r);i++) #define rep1(i,l,r) for(int i = (l);i <= (r);i++) #define MS0(a) memset(a,0,sizeof(a)) #define MS1(a) memset(a,-1,sizeof(a)) double a[11][11],A[12][12]; int n; void Gauss() { int i,j,k; rep1(i,1,n){ int mx = i; rep1(j,i+1,n) if(fabs(A[mx][i]) < fabs(A[j][i])) mx = j; rep1(j,i,n+1) swap(A[mx][j],A[i][j]); rep1(j,i+1,n)if(A[i][i] != 0){ double y = A[j][i]/A[i][i]; rep1(k,i,n+1) A[j][k] -= y*A[i][k]; } } for(int i = n;i >= 1;i--){ rep1(j,i+1,n) A[i][n+1] -= A[i][j] * A[j][n+1]; A[i][n+1] /= A[i][i]; //化为系数为1;保证有解,则A[i][i] != 0; } } int main() { int i,j; scanf("%d",&n); rep1(i,1,n+1) rep1(j,1,n) scanf("%lf",&a[i][j]); rep1(i,1,n) rep1(j,1,n){ A[i][j] = 2*(a[i+1][j] - a[i][j]); A[i][n+1] += a[i+1][j]*a[i+1][j] - a[i][j]*a[i][j]; } Gauss(); printf("%.3f",A[1][n+1]); rep1(i,2,n) printf(" %.3f",A[i][n+1]); }