codeforces2C 模拟退火

比较显然的模拟退火。
如果我们知道了点的坐标，那么其与其他三个圆的角度就可以得到了。我们需要最大化角度，这点不是很清楚，大概是因为这个点唯一?

#include <bits/stdc++.h>using namespace std;struct Circle {    double x, y, r;    void in() {        scanf("%lf%lf%lf", &x, &y, &r);    }} cir[3];double Length(double x1, double y1, double x2, double y2) {    return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));}struct SimulatedAnnealing {    const double eps = 1e-6;    double x, y, dis, t, r, delta;    int best;    double walk[4][2] = {{0, -1}, {1, 0}, {0, 1}, {-1, 0}};    double check(double x, double y) {        double cta[3];        double tmp = 0, tmp1 = 0;        for(int i = 0; i < 3; i++)            cta[i] = Length(cir[i].x, cir[i].y, x, y)/cir[i].r, tmp += cta[i];        tmp /= 3.0;        for(int i = 0; i< 3; i++) tmp1 += (tmp-cta[i])*(tmp-cta[i]);        return tmp1;    }    pair<double, double> gao() {        delta = 1;        x = y = 0;        for(int i = 0; i < 3; i++) cir[i].in(), x += cir[i].x, y += cir[i].y;        x /= 3.0, y /= 3.0;        while(delta > eps) {            r = check(x, y), best = -1;            for(int i = 0; i < 4; i++) {                t = check(x+walk[i][0]*delta, y+walk[i][1]*delta);                if(t < r) {                    r = t;                    best = i;                }            }            if(best == -1) delta *= 0.7;            else x += walk[best][0]*delta, y += walk[best][1]*delta;        }//        cout<<x<<" "<<y<<endl;        return {x, y};    }} solver;int main(){    pair<double, double> p = solver.gao();    if(solver.check(p.first, p.second) < 1e-6) printf("%.10f %.10f", p.first, p.second);    return 0;}