BZOJ1337: 最小圆覆盖

题目：求n个点的最小圆覆盖。

题解：最小圆覆盖，上模板。复杂度证明可以戳：这里

代码：

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#define inf 1000000000
#define maxn 1000000+5
#define maxm 200000+5
#define eps 1e-6
#define ll long long
#define ull unsigned long long
#define pa pair<int,int>
#define for0(i,n) for(int i=0;i<=(n);i++)
#define for1(i,n) for(int i=1;i<=(n);i++)
#define for2(i,x,y) for(int i=(x);i<=(y);i++)
#define for3(i,x,y) for(int i=(x);i>=(y);i--)
#define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
#define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
#define mod 1000000007
#define lch k<<1,l,mid
#define rch k<<1|1,mid+1,r
#define sqr(x) (x)*(x)
#define db double
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
    return x*f;
}
struct point
{
    db x,y;
    point operator +(point b){return (point){x+b.x,y+b.y};}
    point operator /(db b){return (point){x/b,y/b};}
}a[maxn];
int n;
db ans;
inline db dist(point a,point b){return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}
inline point calc(db a,db b,db c,db d,db e,db f)
{
  return (point){(c*e-f*b)/(a*e-d*b),(a*f-c*d)/(a*e-d*b)};
}
inline point get(point a,point b,point c)
{
   return calc(b.x-a.x,b.y-a.y,(sqr(b.x)+sqr(b.y)-sqr(a.x)-sqr(a.y))/2.0,
        c.x-b.x,c.y-b.y,(sqr(c.x)+sqr(c.y)-sqr(b.x)-sqr(b.y))/2.0);
}
int main()
{
   freopen("input.txt","r",stdin);
   freopen("output.txt","w",stdout);
   n=read();
   for1(i,n)scanf("%lf%lf",&a[i].x,&a[i].y);
   for1(i,n)swap(a[rand()%n+1],a[rand()%n+1]);
   ans=0;
   for1(i,n)if(dist(a[i],a[0])>ans+eps)
   {
        a[0]=a[i];ans=0;
        for1(j,i-1)if(dist(a[j],a[0])>ans+eps)
        {
            a[0]=(a[i]+a[j])/2;ans=dist(a[0],a[i]);
            for1(k,j-1)if(dist(a[k],a[0])>ans+eps)
            {
                a[0]=get(a[i],a[j],a[k]);ans=dist(a[0],a[i]);
            }
        }
   }
   printf("%.2f %.2f %.2f
",a[0].x,a[0].y,ans);
   return 0;
}

我不会说告诉你这题是三倍经验的