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 = ;
 const int L = ;
 double X,Y;
 int n;
 double mindis[N];

 struct Point{
     double x , y;
     Point(double x= , double y=):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 = ;
     for(int i= ; 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= ; i<n ; i++) p[i].input();
         for(int i= ; i<P ; i++){
             tmp[i].x = random()/1000.0*X;
             tmp[i].y = random()/1000.0*Y;
             mindis[i] = cal(tmp[i]);
         }
         double step = sqrt(X*X+Y*Y)/;
         while(step>1e-){
             for(int i= ; i<P ; i++){
                 for(int j= ; j<L ; j++){
                     Point cur;
                     double ang = random()/1000.0**PI;
                     cur.x = tmp[i].x+cos(ang)*step;
                     cur.y = tmp[i].y+sin(ang)*step;
                     if(cur.x< || cur.x>X || cur.y< || 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= ; i<P ; i++){
             if(mindis[i]<ret){
                 u = tmp[i];
                 ret = mindis[i];
             }
         }
         printf("(%.1f,%.1f).\n%.1f\n" , u.x,u.y,ret);
     }
     return ;
 }