题意：n个（n<=10000）质量分别为 mi 的质点的坐标分别为 (xi,yi) （数据均为整数），每个重物连着一根绳子，所有绳子有一个公共的绳结，绳结只能在一个与地面平行的平面上移动，质点不会落地，求绳结最终停留的坐标，保留小数点后三位。 ?100000≤xi,yi≤100000 。
可以用模拟退火，更简单的方法是模拟：给绳结一个初始位置，求受力，作用一段时间，产生一段位移。
嗯，都叫模拟……看是想模拟热力学还是力学 QAQ
模拟退火：
http://blog.csdn.net/popoqqq/article/details/39322569
https://zhuanlan.zhihu.com/p/21277465
这个问题其实叫广义费马点：
http://www.matrix67.com/blog/archives/422
#include <cstdio>
#include <cmath>
#include <cstdlib>
using namespace std;
const int MAX_N = 10000, loop = 1000;
const double pre = 1e-3, init = 1e4, c = 0.993, pi = acos(-1), inf = 1e70;
int n;
struct Point {double x, y, w;Point(double x=0, double y=0, double w=0): x(x), y(y), w(w) {}Point operator+(Point b){return Point(x+b.x, y+b.y);}friend Point operator*(double k, Point p){return Point(k*p.x, k*p.y);}
} P[MAX_N], a;
double sum, e = inf;inline double dist(const Point& a, const Point& b)
{return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}inline double J(const Point& p)
{double r = 0;for (int i = 0; i < n; ++i)r += dist(p, P[i]) * P[i].w;if (r < e) {a = p;e = r;}return r;
}inline double r()
{return (double)rand()/RAND_MAX;
}inline void move(Point& p, Point& q, double d)
{double theta = 2*pi*r();q = p + d*Point(cos(theta), sin(theta));
}void SA()
{Point p, q;for (int i = 0; i < n; ++i)p = p + P[i].w/sum*P[i];double last = J(p);for (double T = init; T >= pre; T *= c) {move(p, q, T);double now = J(q), dE = last - J(q);if (dE > 0 || r() < exp(dE/T)) {p = q;last = now;}}for (int i = 0; i < loop; ++i) {move(a, q, pre);J(q);}
}int main()
{srand(2333333);scanf("%d", &n);for (int i = 0; i < n; ++i) {scanf("%lf %lf %lf", &P[i].x, &P[i].y, &P[i].w);sum += P[i].w;}SA();printf("%.3f %.3f\n", a.x, a.y);return 0;
}