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+;
 const double eps= 1e-;
 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 =  * (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=;
     for(int i=;i<=n;i++)
     {
         if(fabs(dis(a[i],mid)-d)<=eps)
             ans++;
         if((ans+(n-i))*<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*>=n)
         return true;
     return false;
 }
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         for(int i=;i<=n;i++)
             scanf("%lf%lf",&a[i].x,&a[i].y);
         if(n<=)                                   //n小于等于4特判
         {
             printf("%lf %lf %lf\n",a[].x,a[].y,0.0);
             continue;
         }
         else if(n<=)
         {
             printf("%lf %lf %lf\n",(a[].x+a[].x)/,(a[].y+a[].y)/,dis(a[],a[])/);
             continue;
         }
         while(true)
         {
             point aa=a[rand()%n+],bb=a[rand()%n+],cc=a[rand()%n+];   //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\n",aa.x,aa.y);
 //                printf("%lf %lf\n",bb.x,bb.y);
 //                printf("%lf %lf\n",cc.x,cc.y);
                 printf("%lf %lf %lf\n",xin.x,xin.y,r);
                 break;
             }
         }
     }
     return ;
 }