2017 CCPC 哈尔滨站 HDU 6242

Geometry Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1091    Accepted Submission(s): 208
Special Judge

Problem Description

Alice is interesting in computation geometry problem recently. She found a interesting problem and solved it easily. Now she will give this problem to you :

You are given N distinct points (Xi,Yi) on the two-dimensional plane. Your task is to find a point P and a real number R, such that for at least ⌈N2⌉ given points, their distance to point P is equal to R.

 

Input

The first line is the number of test cases.

For each test case, the first line contains one positive number N(1≤N≤105).

The following N lines describe the points. Each line contains two real numbers Xi and Yi (0≤|Xi|,|Yi|≤103) indicating one give point. It's guaranteed that Npoints are distinct.

 

Output

For each test case, output a single line with three real numbers XP,YP,R, where (XP,YP) is the coordinate of required point P. Three real numbers you output should satisfy 0≤|XP|,|YP|,R≤109.

It is guaranteed that there exists at least one solution satisfying all conditions. And if there are different solutions, print any one of them. The judge will regard two point's distance as R if it is within an absolute error of 10−3 of R.

 

Sample Input

1

7

1 1

1 0

1 -1

0 1

-1 1

0 -1

-1 0

 

Sample Output

0 0 1

  题意  给出n个点 确定一个圆的圆心和半径  使得至少n/2个点（向上取整）在该圆上 对于每组样例至少有一个解

解析 我们知道  在n个点中每个点在圆上的概率都为0.5  三个不共线的点确定一个外接圆  我们随机取三个点 这三个点的外接圆满足条件的概率为0.5*0.5*0.5=0.125

每次随机消耗的时间复杂度为1e5  枚举1秒内可以100 次   基本可以得到答案

AC代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <queue>
#include <vector>
using namespace std;
const int maxn= 1e5+10;
const double eps= 1e-6;
const int inf = 0x3f3f3f3f;
typedef long long ll;
struct point
{
    double x,y;
}a[maxn];
int n;
double dis(point a,point b)   //两点间距离
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool waijie(point p1,point p2,point p3,point &ans)   //引用修改圆心的值
{
    if(fabs((p3.y-p2.y)*(p2.x-p1.x)-(p2.y-p1.y)*(p3.x-p2.x))<=eps)return false;  //三点共线 没有外接圆
    double Bx = p2.x - p1.x, By = p2.y - p1.y;            //外接圆板子
    double Cx = p3.x - p1.x, Cy = p3.y - p1.y;
    double D = 2 * (Bx * Cy - By * Cx);
    double cx = (Cy * (Bx * Bx + By * By) - By * (Cx * Cx + Cy * Cy)) / D + p1.x;
    double cy = (Bx * (Cx * Cx + Cy * Cy) - Cx * (Bx * Bx + By * By)) / D + p1.y;
    ans.x=cx,ans.y=cy;
    return true;
}
bool check(point mid,double d)     //检查是否有n/2个点在外接圆上
{
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        if(fabs(dis(a[i],mid)-d)<=eps)
            ans++;
        if((ans+(n-i))*2<n)       //简单优化一下 如果还未判断的点的数量加上已经满足条件的点的数量小于n/2 false
            return false;
    }
    if(ans*2>=n)
        return true;
    return false;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%lf%lf",&a[i].x,&a[i].y);
        if(n<=2)                                   //n小于等于4特判
        {
            printf("%lf %lf %lf
",a[1].x,a[1].y,0.0);
            continue;
        }
        else if(n<=4)
        {
            printf("%lf %lf %lf
",(a[1].x+a[2].x)/2,(a[1].y+a[2].y)/2,dis(a[1],a[2])/2);
            continue;
        }
        while(true)
        {
            point aa=a[rand()%n+1],bb=a[rand()%n+1],cc=a[rand()%n+1];   //随机产生3个点
            point xin;
            if(!waijie(aa,bb,cc,xin))
                continue;
            double r=dis(aa,xin);
            if(check(xin,r))
            {
//                printf("%lf %lf
",aa.x,aa.y);
//                printf("%lf %lf
",bb.x,bb.y);
//                printf("%lf %lf
",cc.x,cc.y);
                printf("%lf %lf %lf
",xin.x,xin.y,r);
                break;
            }
        }
    }
    return 0;
}