[bzoj3680] 吊打XXX：模拟退火 or 模拟力学情景

来源：互联网发布：淘宝直播怎么联系商家编辑：程序博客网时间：2024/05/02 00:35

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;}

0 0