hdu 6242 Geometry Problem

Geometry Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1722    Accepted Submission(s): 304
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 N points 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

 

这一题让我学会了随机数法。

 

题意：找出一个圆使至少n/2个点在圆上（数据保证有解）

 

解题思路：用普通的for循环肯定会超，因为数据保证有解，我们随机三个点，这三个点都在圆上的概率是0.5*0.5*0.5=0.125，三个点不都在圆上的概率是7/8，那么随机100次，概率就是(7/8)^100,约为1e-6，基本接近于0了，也就是说随机一百遍基本就能找到三个点都在圆上的情况，也就能找到那个圆的圆心和半径了。

 

注意坑点：1.n<5的情况要另考虑　　2.三点共线　　3.判断找到圆的条件（我一开始是cnt<=n/2,这样n如果从0开始遍历答案就不对了

 

附ac代码：

#include <iostream>
#include <string.h>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
struct nod
{
    double x;
    double y;
}nu[maxn];
int vis[maxn];
double xx1,yy1,xx2,yy2,xx3,yy3;
void getr(double &x,double &y,double &r)
{
    //   printf("%lf %lf
",xx2,xx1);
    double a=2*(xx2-xx1);
    double b=2*(yy2-yy1);
    double c=xx2*xx2-xx1*xx1+yy2*yy2-yy1*yy1;
    double d=2*(xx3-xx2);
    double e=2*(yy3-yy2);
    double f=xx3*xx3-xx2*xx2+yy3*yy3-yy2*yy2;
    x=(b*f-e*c)/(b*d-e*a);
    y=(a*f-d*c)/(a*e-b*d);
    r=sqrt((x-xx1)*(x-xx1)+(y-yy1)*(y-yy1));
    //   printf("%lf %lf %lf
",x,y,r);
}
int main()
{
    int t;
    int n;
    scanf("%d",&t);
    while(t--)
    {
        
        scanf("%d",&n);
        for(int i=0;i<n;++i)
            scanf("%lf%lf",&nu[i].x,&nu[i].y);
        if(n<=2)
        {
            printf("%lf %lf %lf
",nu[0].x,nu[0].y,0.0);
        }
        else if(n<=4)
        {
            double x,y,r;
            x=(nu[0].x+nu[1].x)/2;
            y=(nu[0].y+nu[1].y)/2;
            r=sqrt((x-nu[0].x)*(x-nu[0].x)+(y-nu[0].y)*(y-nu[0].y));
            printf("%lf %lf %lf
",x,y,r);
        }
        else
        {
            while (true)
            {
                int coo1=rand()%n;
                int coo2=rand()%n;
                int coo3=rand()%n;
                if(coo1==coo2 || coo1==coo3 || coo2==coo3) continue;
                xx1=nu[coo1].x;  yy1=nu[coo1].y;
                xx2=nu[coo2].x;  yy2=nu[coo2].y;
                xx3=nu[coo3].x;  yy3=nu[coo3].y;
                if(fabs((yy3-yy2)*(xx2-xx1)-(xx3-xx2)*(yy2-yy1))<=1e-6)
                    continue;
                double x=0,y=0,r=0;
                getr(x,y,r);
                int cnt=0;
                for(int i=0;i<n;++i)
                {
                    if(fabs(r*r- ((nu[i].x-x)*(nu[i].x-x)+(nu[i].y-y)*(nu[i].y-y)) )<=1e-6)
                        ++cnt;
                }
                if(cnt*2>=n)
                {
                    printf("%lf %lf %lf
",x,y,r);
                    break;
                }
            }
        }
    }
    return 0;
}