POJ 2420 A Star not a Tree? 费马点 计算几何 模拟退火

题目大意：给出平面中一些点，求平面中的一个点，使得这个点到其他所有点的距离之和最短，并输出这个距离和。

思路：模拟退火。精度到整数。。。没什么需要注意的。。

话说费马点只能用模拟退火求近似解吧

CODE：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 510
#define INF 0x7f7f7f7f
#define EPS 1e-2
#define PI acos(-1.0)
using namespace std;

struct Point{
	double x,y;
	double length;

	Point(double _x,double _y):x(_x),y(_y) {}
	Point() {}

	void Read() {
		scanf("%lf%lf",&x,&y);
	}
}point[MAX],now,ans;

int points;
double dE;

void Simulated_Annealing();
inline double Rand();
inline double Statistic(Point p);
inline double Calc(Point p1,Point p2);

int main()
{	
	cin >> points;
	for(int i = 1;i <= points; ++i)
		point[i].Read();
	now = Point(5000.0,5000.0);
	min_length = Statistic(now);
	Simulated_Annealing();
	printf("%.0f",ans.length);
	return 0;
}

void Simulated_Annealing()
{
	double L = 10000.0;
	while(L > EPS) {
		double alpha = 2 * PI * Rand();
		Point temp(now.x + cos(alpha) * T,now.y + sin(alpha) * T);
		dE = Statistic(now) - Statistic(temp);
		if(dE >= 0 || exp(dE / L) > Rand)
			now = temp;
		L *= .99;
	}
}

inline double Rand()
{
	return (rand() % 1000.0 + 1) / 1000.0;
}

inline double Statistic(Point p)
{
	double re = 0.0;
	for(int i = 1;i <= points; ++i)
		re += Calc(point[i],p);
	return re;
}

inline double Calc(Point p1,Point p2)
{
	return sqrt((p1.x - p2.x) * (p1.x - p2.x) + 
				(p1.y - p2.y) * (p1.y - p2.y));
}