zoj - 2928 - Mathematical contest in modeling（爬山）

題意：空間中有 n(3 <= n <= 100) 個點(0.00 <= ai, bi, ci <= 1000.00)，求到這 n 個點的距離之和最短的一點。

題目鏈接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2928

——>>如果是一維的，那麼答案是中位數；現在是三維的，賽時不會，賽後學了一種叫做爬山的隨機算法用於這題。

爬山算法：先選取其中一值作爲最優值，然後向該值附近掃描，若發現更優的值，則以該更優值作爲最優值，繼續迭代掃描；否則所選值已是最優值。。

此題我以(0, 0, 0)作爲初始最優值，然後往其27個可前進方向移動尋找更優值。

精度的設置要特別小心。。1e-8的精度過不了。。

#include <cstdio>
#include <cmath>

const int MAXN = 100 + 10;
const int MAXD = 30;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-12;
const double rate = 0.99;

int n;
int dx[MAXD];
int dy[MAXD];
int dz[MAXD];
int dcnt;
double a[MAXN];
double b[MAXN];
double c[MAXN];
double A, B, C;

void Read()
{
    for (int i = 0; i < n; ++i)
    {
        scanf("%lf%lf%lf", a + i, b + i, c + i);
    }
}

void getDirection()
{
    dcnt = 0;
    for (int i = -1; i <= 1; ++i)
    {
        for (int j = -1; j <= 1; ++j)
        {
            for (int k = -1; k <= 1; ++k)
            {
                dx[dcnt] = i;
                dy[dcnt] = j;
                dz[dcnt] = k;
                ++dcnt;
            }
        }
    }
}

int Dcmp(double x)
{
    if (fabs(x) < EPS) return 0;
    return x > 0 ? 1 : -1;
}

void Solve()
{
    double step = 1000;
    double minDistance = INF;

    A = B = C = 0;
    getDirection();
    while (Dcmp(step) != 0)
    {
        for (int i = 0; i < dcnt; ++i)
        {
            double newx = A + dx[i] * step;
            double newy = B + dy[i] * step;
            double newz = C + dz[i] * step;
            double sumOfDistance = 0;

            if (Dcmp(newx) < 0 || Dcmp(newx - 1000) > 0 ||
                Dcmp(newy) < 0 || Dcmp(newy - 1000) > 0 ||
                Dcmp(newz) < 0 || Dcmp(newz - 1000) > 0) continue;

            for (int j = 0; j < n; ++j)
            {
                sumOfDistance += sqrt((a[j] - newx) * (a[j] - newx) + (b[j] - newy) * (b[j] - newy) + (c[j] - newz) * (c[j] - newz));
            }
            if (Dcmp(sumOfDistance - minDistance) < 0)
            {
                minDistance = sumOfDistance;
                A = newx;
                B = newy;
                C = newz;
            }
        }
        step *= rate;
    }
}

void Output()
{
    printf("%.3f %.3f %.3f\n", A, B, C);
}

int main()
{
    while (scanf("%d", &n) == 1)
    {
        Read();
        Solve();
        Output();
    }

    return 0;
}