HDU 3932 模拟退火

HDU3932

 题目大意：给定一堆点，找到一个点的位置使这个点到所有点中的最大距离最小

简单的模拟退火即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <algorithm>

using namespace std;

#define N 1005
#define PI acos(-1.0)
#define random(x) (rand()%x+1)
const int P = 20;
const int L = 25;
double X,Y;
int n;
double mindis[N];

struct Point{
    double x , y;
    Point(double x=0 , double y=0):x(x),y(y){}
    void input(){
        scanf("%lf%lf" , &x , &y);
    }
}p[N] , tmp[N];

double dis(Point a , Point b)
{
    double x = a.x-b.x , y=a.y-b.y;
    return sqrt(x*x+y*y);
}

double cal(Point a)
{
    double maxn = 0;
    for(int i=0 ; i<n ; i++) maxn = max(maxn , dis(a , p[i]));
    return maxn;
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("a.in" , "r" , stdin);
    #endif // ONLINE_JUDGE
    while(~scanf("%lf%lf%d" , &X , &Y , &n))
    {
        for(int i=0 ; i<n ; i++) p[i].input();
        for(int i=0 ; i<P ; i++){
            tmp[i].x = random(1000)/1000.0*X;
            tmp[i].y = random(1000)/1000.0*Y;
            mindis[i] = cal(tmp[i]);
        }
        double step = sqrt(X*X+Y*Y)/2;
        while(step>1e-3){
            for(int i=0 ; i<P ; i++){
                for(int j=0 ; j<L ; j++){
                    Point cur;
                    double ang = random(1000)/1000.0*2*PI;
                    cur.x = tmp[i].x+cos(ang)*step;
                    cur.y = tmp[i].y+sin(ang)*step;
                    if(cur.x<0 || cur.x>X || cur.y<0 || cur.y>Y) continue;
                    double val = cal(cur);
                    if(val<mindis[i]){
                        mindis[i] = val;
                        tmp[i] = cur;
                    }
                }
            }
            step *= 0.85;
        }
        double ret = 1e20;
        Point u;
        for(int i=0 ; i<P ; i++){
            if(mindis[i]<ret){
                u = tmp[i];
                ret = mindis[i];
            }
        }
        printf("(%.1f,%.1f).
%.1f
" , u.x,u.y,ret);
    }
    return 0;
}