LC 1515. Best Position for a Service Centre (Simulated Annealing)

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class Solution {
public:
    int n;
    double eps=1E-6;
    const int dx[4]={-1,0,1,0};
    const int dy[4]={0,1,0,-1};
    double getMinDistSum(vector<vector<int>>& positions) {
        n=positions.size();
        double xavg=0.0;
        double yavg=0.0;
        for(auto& p:positions){
            xavg+=p[0];
            yavg+=p[1];
        }
        xavg/=n;
        yavg/=n;
        double mindis=dist(xavg,yavg,positions);
        double step=100.0;
        int done=0;
        while(step>eps){
            done=0;
            for(int i=0;i<4;i++){
                double nx=xavg+step*dx[i];
                double ny=yavg+step*dy[i];
                
                double t=dist(nx,ny,positions);
                if(t<mindis){
                    mindis=t;
                    xavg=nx;
                    yavg=ny;
                    done=1;
                    break;
                }
            }
            if(!done){
                step/=2;
            }
        }
        return mindis;
    }
    
    double dist(double x, double y,vector<vector<int>>& positions){
        double ret=0.0;
        for(int i=0;i<n;i++){
            double dx=positions[i][0]-x;
            double dy=positions[i][1]-y;
            ret+=sqrt(dx*dx+dy*dy);
        }
        return ret;
    }
};