hdu 3932 最小覆盖圆 随机算法做法
找出一个点使得这个店到n个点的最长距离最短,即求最小覆盖圆的半径
用一个点往各个方向扩展,如果结果更优,则继续以当前步长扩展,否则缩小步长
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<stdio.h>
#include<math.h>
#include<string.h>
const double pi = acos(-1.0);
struct point {
double x,y;
}p[1010];
int n;
point mid,tmid;
double R,xmin,ymin,xmax,ymax;
double ans,tans;
double dis(point a, point b)
{
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
void solve(double x,double y)
{
double max=0;
point tp;
tp.x=x;
tp.y=y;
for(int i=0;i<n;i++)
{
double tmp=dis(p[i],tp);
if(tmp>max)
max=tmp;
}
if(max<tans)
{
tans=max;
tmid=tp;
}
}
int main()
{
int X,Y,i,k;
double j;
while(scanf("%d%d%d",&X,&Y,&n)!=EOF)
{
// printf("n=%d\n",n);
xmin=10000;ymin=10000;xmax=0;ymax=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
if(p[i].x<xmin) xmin=p[i].x;
if(p[i].y<xmin) xmin=p[i].y;
if(p[i].x>xmax) xmax=p[i].x;
if(p[i].y>ymax) ymax=p[i].y;
}
mid.x=(xmin+xmax)/2;
mid.y=(ymin+ymax)/2;
tmid=mid;
R=sqrt((xmax-xmin)*(xmax-xmin)+(ymax-ymin)*(ymax-ymin))/2;
for(i=0;i<n;i++)
{
double tmp=dis(mid,p[i]);
if(tmp>tans)
tans=tmp;
}
ans=tans;
while(1)
{
for(double j=0;j<=2*pi;j+=0.1)
{
solve(mid.x+R*sin(j),mid.y+R*cos(j));
}
if(fabs(ans-tans)<0.01&&R<0.0001) break;
if(tans<ans)
{
ans=tans;
mid=tmid;
}
else R*=0.7;
}
printf("(%.1lf,%.1lf).\n", mid.x, mid.y);
printf("%.1lf\n", sqrt(ans));
}
return 0;
}