zoj 1450 Minimal Circle（最小包围圆）

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本文介绍了一种寻找能够覆盖一组点且面积最小的圆的方法。通过随机化算法和计算几何原理，实现了一个模板题的解决方案，并附带了完整的C++代码实现。

Minimal Circle

Time Limit: 5 Seconds Memory Limit: 32768 KB

You are to write a program to find a circle which covers a set of points and has the minimal area. There will be no more than 100 points in one problem.

Input

The input contains several problems. The first line of each problem is a line containing only one integer N which indicates the number of points to be covered. The next N lines contain N points. Each point is represented by x and y coordinates separated by a space. After the last problem, there will be a line contains only a zero.

Output

For each input problem, you should give a one-line answer which contains three numbers separated by spaces. The first two numbers indicate the x and y coordinates of the result circle, and the third number is the radius of the circle. (use escape sequence %.2f)

Sample Input

2
0.0 0.0
3 0
5
0 0
0 1
1 0
1 1
2 2
0

Sample Output

1.50 0.00 1.50
1.00 1.00 1.41

Source: Asia 1997, Shanghai (Mainland China)

题目：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1450

题意：给你n个点，求一个最小的圆，能够包围这些点

分析：这题直接是模板题，不过我不会啊，最后找了好多资料，发现《计算几何--算法与应用》第4.7章里面讲得不错，也有证明。。。

模板直接抄袭某牛的= =

代码：

#include<cmath>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int mm=111;
typedef double diy;
struct point
{
    diy x,y;
}g[mm];
diy Sqr(diy x)
{
    return x*x;
}
diy Dis(point P,point Q)
{
    return sqrt(Sqr(P.x-Q.x)+Sqr(P.y-Q.y));
}
void Circle(point P0,point P1,point P2,point &o)
{
    diy a1=P1.x-P0.x,b1=P1.y-P0.y,c1=(Sqr(a1)+Sqr(b1))/2;
    diy a2=P2.x-P0.x,b2=P2.y-P0.y,c2=(Sqr(a2)+Sqr(b2))/2;
    diy d=a1*b2-a2*b1;
    o.x=P0.x+(c1*b2-c2*b1)/d;
    o.y=P0.y+(a1*c2-a2*c1)/d;
}
void MinCircle(point g[],point &o,diy &r,int n)
{
    random_shuffle(g,g+n);
    int i,j,k;
    o=g[0];
    for(r=0,i=1;i<n;++i)
    {
        if(Dis(g[i],o)<=r)continue;
        o=g[i];
        for(r=j=0;j<i;++j)
        {
            if(Dis(g[j],o)<=r)continue;
            o.x=(g[i].x+g[j].x)/2;
            o.y=(g[i].y+g[j].y)/2;
            r=Dis(o,g[i]);
            for(k=0;k<j;++k)
            {
                if(Dis(g[k],o)<r)continue;
                Circle(g[i],g[j],g[k],o);
                r=Dis(o,g[i]);
            }
        }
    }
}
int main()
{
    int i,n;
    point o;
    diy r;
    while(scanf("%d",&n),n)
    {
        for(i=0;i<n;++i)
            scanf("%lf%lf",&g[i].x,&g[i].y);
        MinCircle(g,o,r,n);
        printf("%.2lf %.2lf %.2lf\n",o.x,o.y,r);
    }
    return 0;
}