[洛谷P4035][JSOI2008]球形空间产生器

题目大意：给你$n$个点坐标，要你求出圆心

题解：随机化，可以随机一个点当圆心，然后和每个点比较，求出平均距离$r$，如果到这个点的距离大于$r$，说明离这个点远了，就给圆心施加一个向这个点的力；若小于$r$，说明近了，就施加一个远离这个点的力。所有点比较完后，把假设的圆心按合力方向移动一个距离，距离和当前温度有关。时间越久，温度越低

卡点：第$8$个点精度总是不够，拼命调参，调好后第$3$个点就$Tle$了，最后卡时过的

C++ Code：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#define maxn 12
int n;
struct Point {
	double x[maxn], w;
	inline friend Point operator + (const Point &lhs, const Point &rhs) {
		Point res;
		for (register int i = 0; i < n; i++) res.x[i] = lhs.x[i] + rhs.x[i];
		return res;
	}
	inline friend Point operator - (const Point &lhs, const Point &rhs) {
		Point res;
		for (register int i = 0; i < n; i++) res.x[i] = lhs.x[i] - rhs.x[i];
		return res;
	}
	inline void operator /= (const int &rhs) {
		for (register int i = 0; i < n; i++) x[i] /= rhs;
	}
} s[maxn], ans;

inline double sqr(double x) {return x * x;}
inline double abs(const Point &O) {
	double res = 0;
	for (int i = 0; i < n; i++) res += sqr(O.x[i]);
	return sqrt(res);
}
inline double dis(Point O) {
	double res = 0;
	for (int i = 0; i <= n; i++) res += abs(O - s[i]);
	return res;
}

const double ST = 5000, delT = 0.99995, eps = 1e-5;
const int Tim = 1, __Tim = 250000;
double V[maxn];
void SA() {
	double T = ST;
	while (T > eps) {
		double sum = dis(ans) / (n + 1);
		for (int i = 0; i < n; i++) {
			V[i] = 0;
			for (int j = 0; j <= n; j++) {
				V[i] += (abs(ans - s[j]) - sum) * (s[j].x[i] - ans.x[i]);
			}
			V[i] /= n + 1;
		}
		for (int i = 0; i < n; i++) {
			ans.x[i] += T * V[i];
		}
		T *= delT;
	}
	for (int Tim = 0; Tim < __Tim && 1. * clock() / CLOCKS_PER_SEC < .9; Tim++) {
		double sum = dis(ans) / (n + 1);
		for (int i = 0; i < n; i++) {
			V[i] = 0;
			for (int j = 0; j <= n; j++) {
				V[i] += (abs(ans - s[j]) - sum) * (s[j].x[i] - ans.x[i]);
			}
			V[i] /= n + 1;
		}
		for (int i = 0; i < n; i++) {
			ans.x[i] += T * V[i];
		}
	}
}

int main() {
	scanf("%d", &n);
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j < n; j++) scanf("%lf", s[i].x + j);
		ans = ans + s[i];
	}
	ans /= n + 1;
	for (int i = 0; i < Tim; i++) SA();
	for (int i = 0; i < n; i++) {
		 printf("%.3lf", ans.x[i]);
		 putchar(i == n - 1 ? '\n' : ' ');
	}
	return 0;
}