bzoj1337 最小圆覆盖 计算几何 解题报告

Description

给出平面上N个点,N<=10^5.请求出一个半径最小的圆覆盖住所有的点

Input

第一行给出数字N，现在N行，每行两个实数x,y表示其坐标.

Output

输出最小半径，输出保留三位小数.

Sample Input

4
1 0
0 1
0 -1
-1 0

Sample Output

1.000

思路

数据太弱了吧。。。。

代码

借鉴朱爷

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define N 100010struct Point{      double x,y;             Point(double _ = .0,double __ = .0):x(_),y(__) {}      Point operator +(const Point &a)     {return Point(x+a.x,y+a.y);}      Point operator -(const Point &a)    {return Point(x-a.x,y-a.y);}      Point operator *(double a)    {return Point(x*a,y*a);}      double operator * (const Point &a)    {return x*a.x+y*a.y;}    double operator ^ (const Point &a)    {return x*a.y-y*a.x;}}p[N];struct Line{    Point p,v;    Line(){}    Line(Point _a,Point _b):p(_a),v(_b){}};struct Circle{    Point p;double R;    Circle(){}    Circle(Point _p,double _R):p(_p),R(_R){}}ans;double dis(Point a,Point b) {return sqrt((a-b)*(a-b));}bool in_cir(Point a,Circle b) {return dis(a,b.p)<=b.R;}Point get_intersection(Line &l1,Line &l2){    Point u=l1.p-l2.p;    double tmp=(l2.v^u)/(l1.v^l2.v);    return l1.p+l1.v*tmp; }Point rotate (Point a) {return Point(-a.y,a.x);}int n;int main(){//  srand(19980402);     scanf("%d",&n);    for(int i=1;i<=n;++i) scanf("%lf%lf",&p[i].x,&p[i].y);//  random_shuffle(p+1,p+n+1);    for(int i=1;i<=n;++i)    {        if(!in_cir(p[i],ans))        {            ans.p=p[i];            for(int j=1;j<i;++j)            {                if(!in_cir(p[j],ans))                {                    ans.p=(p[i]+p[j])*0.5;                    ans.R=dis(p[i],p[j])*0.5;                    for(int k=1;k<j;k++)                    {                        if(!in_cir(p[k],ans))                        {                            Line l1((p[i]+p[j])*0.5,rotate(p[i]-p[j]));                            Line l2((p[i]+p[k])*0.5,rotate(p[i]-p[k]));                            ans.p=get_intersection(l1,l2);                            ans.R=dis(ans.p,p[i]);                        }                     }                }            }        }    }    printf("%.3f",ans.R);}