【BZOJ-1336&1337】Alie最小圆覆盖最小圆覆盖（随机增量法）

1336: [Balkan2002]Alien最小圆覆盖

Time Limit: 1 Sec Memory Limit: 162 MBSec Special Judge
Submit: 1573 Solved: 697
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Description

给出N个点，让你画一个最小的包含所有点的圆。

Input

先给出点的个数N,2<=N<=100000，再给出坐标Xi,Yi.(-10000.0<=xi,yi<=10000.0)

Output

输出圆的半径，及圆心的坐标

Sample Input

6
8.0 9.0
4.0 7.5
1.0 2.0
5.1 8.7
9.0 2.0
4.5 1.0

Sample Output

5.00
5.00 5.00

HINT

Source

1337: 最小圆覆盖

Time Limit: 1 Sec Memory Limit: 64 MB
Submit: 897 Solved: 437
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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

HINT

Source

Solution

最小圆覆盖裸题，随机增量法

这道题有个需要注意的地方，输出的时候不要只输出2位小数，可能会WA，可以考虑直接输出

直接把课件黏上来= =

Code

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cstring>

#include<cmath>

#include<vector>

#include<queue>

#include<map>

#include<set>

#include<stack>

#include<cstdlib>

#include<string>

#include<bitset>

#include<iomanip>

#define INF 1000000000

#define fi first

#define se second

#define N 100005

#define MP(x,y) make_pair(x,y)

using namespace std;

typedef long long LL;

typedef pair<int,int> pii;

typedef long double Double;

struct Vector

{

    double x,y;

    Vector(double X=,double Y=) {x=X,y=Y;}

};

typedef Vector Point;

typedef vector<Point> Polygon;

const double eps=1e-;

const double pi=acos(-1.0);

struct Line

{

    Point P;

    Vector v;

    double ang;

    Line() {}

    Line(Point P,Vector v):P(P),v(v) {ang=atan2(v.y,v.x);}

    bool operator<(const Line &L) const {return ang<L.ang;}

};

int dcmp(double x) {if(fabs(x)<eps) return ; else return x<? -:;}

Vector operator + (Vector A,Vector B) {return ((Vector){A.x+B.x,A.y+B.y});}

Vector operator - (Vector A,Vector B) {return ((Vector){A.x-B.x,A.y-B.y});}

Vector operator * (Vector A,double p) {return ((Vector){A.x*p,A.y*p});}

Vector operator / (Vector A,double p) {return ((Vector){A.x/p,A.y/p});}

bool operator < (const Vector& a,const Vector& b) {return a.x<b.x||(a.x==b.x&&a.y<b.y);}

bool operator == (const Vector& a,const Vector& b) {return dcmp(a.x-b.x)==&&dcmp(a.y-b.y)==;}

double Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}

double Len(Vector A) {return sqrt(Dot(A,A));}

double Cross(Vector A,Vector B) {return A.x*B.y-A.y*B.x;}

Vector Rotate(Vector A,double rad) {return ((Vector){A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad)});}

Point GLI(Point P,Vector v,Point Q,Vector w) {Vector u=P-Q; double t=Cross(w,u)/Cross(v,w); return P+v*t;}

Point GLI(Line a,Line b) {Vector u=a.P-b.P; double t=Cross(b.v,u)/Cross(a.v,b.v);return a.P+a.v*t;}

Point Center_of_gravity(Point A,Point B,Point C)

{

    Point P=(A+B)/,Q=(A+C)/;

    Vector v=Rotate(B-A,pi/),w=Rotate(C-A,pi/);

    if(dcmp(Len(Cross(v,w)))==)//这是三点一线的情况

    {

        if(dcmp(Len(A-B)+Len(B-C)-Len(A-C))==)

            return (A+C)/;

        if(dcmp(Len(A-C)+Len(B-C)-Len(A-B))==)

            return (A+B)/;

        if(dcmp(Len(A-B)+Len(A-C)-Len(B-C))==)

            return (B+C)/;

    }

    return GLI(P,v,Q,w);

}

double Min_Cover_Circle(Point *p,int n,Point &c)

{

    random_shuffle(p,p+n);

    c=p[];

    double r=;

    int i,j,k;

    for(i=;i<n;i++)

        if(dcmp(Len(c-p[i])-r)>)

        {

            c=p[i],r=;

            for(j=;j<i;j++)

                if(dcmp(Len(c-p[j])-r)>)

                {

                    c=(p[i]+p[j])/;

                    r=Len(c-p[i]);

                    for(k=;k<j;k++)

                        if(dcmp(Len(c-p[k])-r)>)

                        {

                            c=Center_of_gravity(p[i],p[j],p[k]);

                            r=Len(c-p[i]);

                        }

                }

        }

    return r;

}

#define MAXN 100010

Point Po[MAXN];

int main()

{

    srand();

    int n; scanf("%d",&n);

    for (int i=; i<=n; i++)

        {

            double x,y; scanf("%lf%lf",&x,&y);

            Po[i-]=Point(x,y);

        }

    Point c;

    printf("%lf\n",Min_Cover_Circle(Po,n,c));

    printf("%lf %lf",c.x,c.y);

    return ;

}