poj1379 模拟退火

题意：和上题一样。。。就是把最小值换成了最大值。。

 

ref：http://www.cppblog.com/RyanWang/archive/2010/01/21/106112.html

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
using namespace std;

#define eps 1e-3
#define pi  acos(-1.0)
#define POI 15            //独立跑POI次，找最值        tp[1..POI]是随机的初值
#define RUN 40            //迭代次数，本题中即点(tx,ty)向RUN个方向发散
#define INF 99999.999
int X,Y,N,T;
double ans;
int ansi;
struct
{
    double x,y;
}tp[1010],hol[1010];
double sol[1010];

double dist(double x1,double y1,double x2,double y2)
{
    return(sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));
}

double dis(double x,double y)
{
    double tmp=INF;
    for(int i=1;i<=N;i++)
    {
        double tx=hol[i].x,ty=hol[i].y;
        tmp=min(tmp,dist(tx,ty,x,y));
    }
    return tmp;
}

void sa()
{
    for(int i=1;i<=POI;i++)
    {
        tp[i].x=(rand()%1000+1)/1000.0*X;
        tp[i].y=(rand()%1000+1)/1000.0*Y;
        sol[i]=dis(tp[i].x,tp[i].y);
        //printf("%.1f~%.1f=%.1f\n",tp[i].x,tp[i].y,sol[i]);
    }

    double step=1.0*max(X,Y)/sqrt(1.0*N);
    while(step>eps)
    {
        for(int i=1;i<=POI;i++)
        {
            double kx=tp[i].x,ky=tp[i].y;
            double tx=kx,ty=ky;
            for(int j=0;j<RUN;j++)
            {
                double angle=(rand()%1000+1)/1000.0*10*pi;
                kx=tx+cos(angle)*step;
                ky=ty+sin(angle)*step;
                if((kx>X)||(ky>Y)||(kx<0)||(ky<0))    continue;
                double tmp=dis(kx,ky);
                if(tmp>sol[i])
                {
                    tp[i].x=kx;    tp[i].y=ky;
                    sol[i]=tmp;
                }
            }
        }
        step*=0.80;
    }
}


int main()
{
    srand(time(NULL));
    cin>>T;
    //cout<<T<<endl;
    while(T--)
    {
        cin>>X>>Y>>N;
        for(int i=1;i<=N;i++)
            cin>>hol[i].x>>hol[i].y;

        sa();

        ans=0.000;    
        for(int i=1;i<=POI;i++)
        {
            //printf("AA: %.1f~%.1f=%.1f\n",tp[i].x,tp[i].y,sol[i]);
            if(sol[i]>ans)
            {
                ans=sol[i];
                ansi=i;
            }
        }
        printf("The safest point is (%.1f, %.1f).\n",tp[ansi].x,tp[ansi].y);
        //printf("%.1lf\n",ans);
    }
    return 0;
}