BZOJ 3680 吊打XXX 计算几何 模拟退火 广义费马点

题目大意：有个人（gty）被吊打，他机智的使用了分身，但是每个分身有他的重力，把这些gty的分身绑起来，经过一个公共的绳结，求这个绳结最后在哪里。

思路：其实这个题就转化成了：定义一个点到一个分身的距离是两点间的距离 * 分身的重力。求平面内到这些点的距离的和的最小值。和poj2420差不多，这个题只需要在统计的时候吧权值乘上每个分身的重量就可以了。

值得一提的是这个题要求精度到1e-3，写的时候还是有点怕的，但是其实精度把握的合适还是能切这个题的。

一开始MAX开了10000，怎么改参数都是挂。。血的教训啊。。。

CODE：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 10010#define INF 1e17#define EPS 1e-3#define PI acos(-1.0)using namespace std; struct Point{    double x,y;    double weight;    double total_weight;         Point(double _x,double _y):x(_x),y(_y) {}    Point() {}    void Read() {        scanf("%lf%lf%lf",&x,&y,&weight);    }}point[MAX],now,ans; int points; inline double Calc(Point p1,Point p2);inline double Statistic(Point p);inline double Rand();void SinulatedAnnealing(); int main(){    srand(19980531);    cin >> points;    for(int i = 1;i <= points; ++i) {        point[i].Read();        now.x += point[i].x;        now.y += point[i].y;    }    now.x /= points,now.y /= points;    ans.total_weight = INF;    SinulatedAnnealing();    printf("%.3lf %.3lf\n",ans.x,ans.y);    return 0;} 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));} inline double Statistic(Point p){    double re = 0.0;    for(int i = 1;i <= points; ++i)        re += Calc(p,point[i]) * point[i].weight;    if(re < ans.total_weight)        ans = p,ans.total_weight = re;    return re;} void SinulatedAnnealing(){    double T = 100000.0;    while(T > EPS) {        double alpha = 2.0 * PI * Rand();        Point temp(now.x + T * cos(alpha),now.y + T * sin(alpha));        double dE = Statistic(now) - Statistic(temp);        if(dE >= 0 || exp(dE / T) >= Rand())            now = temp;        T *= .99;    }       T = .001;    for(int i = 1;i <= 1000; ++i) {        double alpha = 2.0 * PI * Rand();        Point temp(ans.x + T * cos(alpha) * Rand(),ans.y + T * sin(alpha) * Rand());        Statistic(temp);    }} inline double Rand(){    return (rand() % 1000 + 1) / 1000.0;}