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));}