2018.07.04 BZOJ1336&&1337: Balkan2002Alien最小圆覆盖

1336: [Balkan2002]Alien最小圆覆盖

1337: 最小圆覆盖

Time Limit: 1 Sec Memory Limit: 162 MBSec Special Judge

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

这两道题是最小圆覆盖的裸题，这里主要讲讲求最小圆覆盖的思路。

最小圆覆盖问题就是让你求一个最小的圆使得它能够刚好覆盖掉给出的点。本蒟蒻学习的是一种期望效率O（n）" role="presentation" style="position: relative;">O（n）O（n）的算法。

那么我们怎么做呢？

首先，我们假设前i−1" role="presentation" style="position: relative;">i−1i−1个点已经被现在的圆给覆盖掉了，现在正在处理第i" role="presentation" style="position: relative;">ii个点，那么我们check" role="presentation" style="position: relative;">checkcheck一波，如果当前点在圆外，我们就重构覆盖前i" role="presentation" style="position: relative;">ii个点的最小覆盖圆，具体操作就是先以i" role="presentation" style="position: relative;">ii为圆心作圆，然后从前i−1" role="presentation" style="position: relative;">i−1i−1个点中找一个点j" role="presentation" style="position: relative;">jj使得j" role="presentation" style="position: relative;">jj在圆外，此时最小覆盖圆应该为i,j" role="presentation" style="position: relative;">i,ji,j中点，然后我们再从前j−1" role="presentation" style="position: relative;">j−1j−1个点中找一个点k" role="presentation" style="position: relative;">kk使得k" role="presentation" style="position: relative;">kk在圆外，这时最小覆盖圆自然应该是三角形ijk" role="presentation" style="position: relative;">ijkijk的外接圆。我们一直这样操作到前i" role="presentation" style="position: relative;">ii个点都能够check" role="presentation" style="position: relative;">checkcheck成功为止。

但这样的话最坏情况不应该是O（n3）" role="presentation" style="position: relative;">O（n3）O（n3）吗？是的，因此我们要在最开始的时候将点随机打乱，这样的话期望的效率就应该是O（n）" role="presentation" style="position: relative;">O（n）O（n）的了。还有就是注意T1336" role="presentation" style="position: relative;">T1336T1336应该保留10" role="presentation" style="position: relative;">1010位小数而不是题目上说的2" role="presentation" style="position: relative;">22位。

T1336" role="presentation" style="position: relative;">T1336T1336代码：

#include<bits/stdc++.h>
#define N 100005
using namespace std;
struct point{double x,y;}p[N],O;
struct line{double a,b,c;};
double r;
inline double mul(double x){return x*x;}
inline double dis(point x,point y){return sqrt(mul(x.x-y.x)+mul(x.y-y.y));}
inline bool in_check(point x){return dis(O,x)<=r;}
inline point calc(line a,line b){return point{(b.c*a.b-a.c*b.b)/(a.a*b.b-a.b*b.a),(b.c*a.a-a.c*b.a)/(a.b*b.a-b.b*a.a)};}
int main(){
    int n;
    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);
    r=0;
    for(int i=1;i<=n;++i){
        if(in_check(p[i]))continue;
        O.x=p[i].x,O.y=p[i].y,r=0;
        for(int j=1;j<i;++j){
            if(in_check(p[j]))continue;
            O.x=(p[i].x+p[j].x)/2,O.y=(p[i].y+p[j].y)/2;
            r=dis(O,p[i]);
            for(int k=1;k<j;++k){
                if(in_check(p[k]))continue;
                O=calc(line{2*(p[i].x-p[k].x),2*(p[i].y-p[k].y),mul(p[k].x)+mul(p[k].y)-mul(p[i].x)-mul(p[i].y)},line{2*(p[i].x-p[j].x),2*(p[i].y-p[j].y),mul(p[j].x)+mul(p[j].y)-mul(p[i].x)-mul(p[i].y)});
                r=dis(p[i],O);
            }
        }
    }
    printf("%.10lf\n%.10lf %.10lf",r,O.x,O.y);
    return 0;
}

T1337" role="presentation" style="position: relative;">T1337T1337代码：

#include<bits/stdc++.h>
#define N 100005
using namespace std;
struct point{double x,y;}p[N],O;
struct line{double a,b,c;};
double r;
inline double mul(double x){return x*x;}
inline double dis(point x,point y){return sqrt(mul(x.x-y.x)+mul(x.y-y.y));}
inline bool in_check(point x){return dis(O,x)<=r;}
inline point calc(line a,line b){return point{(b.c*a.b-a.c*b.b)/(a.a*b.b-a.b*b.a),(b.c*a.a-a.c*b.a)/(a.b*b.a-b.b*a.a)};}
int main(){
    int n;
    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);
    r=0;
    for(int i=1;i<=n;++i){
        if(in_check(p[i]))continue;
        O.x=p[i].x,O.y=p[i].y,r=0;
        for(int j=1;j<i;++j){
            if(in_check(p[j]))continue;
            O.x=(p[i].x+p[j].x)/2,O.y=(p[i].y+p[j].y)/2;
            r=dis(O,p[i]);
            for(int k=1;k<j;++k){
                if(in_check(p[k]))continue;
                O=calc(line{2*(p[i].x-p[k].x),2*(p[i].y-p[k].y),mul(p[k].x)+mul(p[k].y)-mul(p[i].x)-mul(p[i].y)},line{2*(p[i].x-p[j].x),2*(p[i].y-p[j].y),mul(p[j].x)+mul(p[j].y)-mul(p[i].x)-mul(p[i].y)});
                r=dis(p[i],O);
            }
        }
    }
    printf("%.3lf",r);
    return 0;
}