[bzoj1336] [Balkan2002]Alien最小圆覆盖

　　最小圆覆盖。。三个for是O(n)的QAQ。。因为随机化后新的点不在当前圆内的几率不大。。

　　学习了下求中垂线的姿势...

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<iostream>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #define d double
 7 using namespace std;
 8 const int maxn=1e5+233;
 9 const d eps=1e-8;
10 struct poi{d x,y;}a[maxn];
11 struct cir{poi O;d r;}now;
12 poi operator +(poi a,poi b){return (poi){a.x+b.x,a.y+b.y};}
13 poi operator -(poi a,poi b){return (poi){a.x-b.x,a.y-b.y};}
14 poi operator /(poi a,d b){return (poi){a.x/b,a.y/b};}
15 int i,j,k,n,m;
16
17 inline d dis(poi a,poi b){
18     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
19 }
20 inline poi rev(poi a){return (poi){a.y,-a.x};}
21 inline poi center(poi a,poi b,poi c){
22     if(fabs((b.y-a.y)*(c.x-a.x)-(c.y-a.y)*(b.x-a.x))<eps){
23         if(a.x>b.x)swap(a,b);
24         if(a.x>c.x)swap(a,c);
25         if(b.x>c.x)swap(b,c);
26         return (a+c)/2.0;
27     }
28     poi x1=(a+b)/2.0,x2=x1+rev(a-b),y1=(a+c)/2.0,y2=y1+rev(a-c);
29     if(fabs(y1.x-y2.x)<eps)swap(x1,y1),swap(x2,y2);
30     d k2=(y1.y-y2.y)/(y1.x-y2.x),b2=y2.y-y2.x*k2;
31     if(fabs(x1.x-x2.x)<eps)return (poi){x1.x,k2*x1.x+b2};
32     d k1=(x1.y-x2.y)/(x1.x-x2.x),b1=x2.y-x2.x*k1;
33     d x=(b2-b1)/(k1-k2);
34     return (poi){x,k1*x+b1};
35 }
36 inline bool oncir(poi a){
37     return dis(a,now.O)<now.r+eps;
38 }
39
40 int main(){
41     scanf("%d",&n);srand(n+233);
42     for(i=1;i<=n;i++)scanf("%lf%lf",&a[i].x,&a[i].y);
43     random_shuffle(a+1,a+n+1);
44     now.O=(a[1]+a[2])/2.0,now.r=dis(now.O,a[1]);
45     for(i=3;i<=n;i++)if(!oncir(a[i])){
46         now.O=(a[1]+a[i])/2.0,now.r=dis(a[i],a[1])/2.0;
47         for(j=2;j<i;j++)if(!oncir(a[j])){
48             now.O=(a[i]+a[j])/2.0,now.r=dis(a[i],a[j])/2.0;
49             for(k=1;k<j;k++)if(!oncir(a[k])){
50                 now.O=center(a[i],a[j],a[k]);
51                 now.r=dis(now.O,a[k]);
52             }
53         }
54     }
55     printf("%.6lf\n",now.r);
56     printf("%.6lf %.6lf\n",now.O.x,now.O.y);
57     return 0;
58 }