学步园

题意：给出一个点到12个点的距离，求这个点的坐标。（坐标维数：11）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4835

——>>因为一定有解，所以转化为线性方程组后直接高斯消元。（训练时我们队FA～激动）

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>

using namespace std;

const int maxn = 13;
typedef double Matrix[maxn][maxn];

void gause_elimination(Matrix A, int n)
{
    int i, j, k, r;
    for(i = 0; i < n; i++)
    {
        r = i;
        for(j = i+1; j < n; j++)
            if(fabs(A[j][i]) > fabs(A[r][i]))
                r = j;
        if(r != i) for(j = 0; j <= n; j++) swap(A[r][j], A[i][j]);

        for(k = i+1; k < n; k++)
        {
            for(j = n; j >= i; j--)
                A[k][j] -= A[k][i] / A[i][i] * A[i][j];
        }
    }

    for(i = n-1; i >= 0; i--)
    {
        for(j = i+1; j < n; j++)
            A[i][n] -= A[j][n] * A[i][j];
        A[i][n] /= A[i][i];
    }
}

int main()
{
    int T, i, j, k;
    scanf("%d", &T);
    Matrix a, A;
    while(T--)
    {
        for(i = 0; i < 12; i++)
            for(j = 0; j < 12; j++)
                scanf("%lf", &a[i][j]);

        memset(A, 0, sizeof(A));
        for(i = 0; i < 11; i++)
        {
            for(j = 0; j < 11 ; j++)
            {
                A[i][j] = a[i][j] - a[i+1][j];
            }

            A[i][11] = a[i+1][11] * a[i+1][11] - a[i][11] * a[i][11];
            for(k = 0; k < 11; k++)
                A[i][11] += a[i][k] * a[i][k] - a[i+1][k] * a[i+1][k];
            A[i][11] /= 2;
        }

        gause_elimination(A, 11);

        for(i = 0;i < 10; i++)
            printf("%.2lf ", A[i][11]);
        printf("%.2lf\n", A[10][11]);
    }
    return 0;
}