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：

 1 #include <iostream>
 2 #include <string.h>
 3 #include <algorithm>
 4 #include <cstdio>
 5 #include <cstdlib>
 6 #include <cmath>
 7 using namespace std;
 8 typedef long long ll;
 9 const int maxn = 1e5+10;
10 struct nod
11 {
12     double x;
13     double y;
14 }nu[maxn];
15 int vis[maxn];
16 double xx1,yy1,xx2,yy2,xx3,yy3;
17 void getr(double &x,double &y,double &r)
18 {
19     //   printf("%lf %lf\n",xx2,xx1);
20     double a=2*(xx2-xx1);
21     double b=2*(yy2-yy1);
22     double c=xx2*xx2-xx1*xx1+yy2*yy2-yy1*yy1;
23     double d=2*(xx3-xx2);
24     double e=2*(yy3-yy2);
25     double f=xx3*xx3-xx2*xx2+yy3*yy3-yy2*yy2;
26     x=(b*f-e*c)/(b*d-e*a);
27     y=(a*f-d*c)/(a*e-b*d);
28     r=sqrt((x-xx1)*(x-xx1)+(y-yy1)*(y-yy1));
29     //   printf("%lf %lf %lf\n",x,y,r);
30 }
31 int main()
32 {
33     int t;
34     int n;
35     scanf("%d",&t);
36     while(t--)
37     {
38
39         scanf("%d",&n);
40         for(int i=0;i<n;++i)
41             scanf("%lf%lf",&nu[i].x,&nu[i].y);
42         if(n<=2)
43         {
44             printf("%lf %lf %lf\n",nu[0].x,nu[0].y,0.0);
45         }
46         else if(n<=4)
47         {
48             double x,y,r;
49             x=(nu[0].x+nu[1].x)/2;
50             y=(nu[0].y+nu[1].y)/2;
51             r=sqrt((x-nu[0].x)*(x-nu[0].x)+(y-nu[0].y)*(y-nu[0].y));
52             printf("%lf %lf %lf\n",x,y,r);
53         }
54         else
55         {
56             while (true)
57             {
58                 int coo1=rand()%n;
59                 int coo2=rand()%n;
60                 int coo3=rand()%n;
61                 if(coo1==coo2 || coo1==coo3 || coo2==coo3) continue;
62                 xx1=nu[coo1].x;  yy1=nu[coo1].y;
63                 xx2=nu[coo2].x;  yy2=nu[coo2].y;
64                 xx3=nu[coo3].x;  yy3=nu[coo3].y;
65                 if(fabs((yy3-yy2)*(xx2-xx1)-(xx3-xx2)*(yy2-yy1))<=1e-6)
66                     continue;
67                 double x=0,y=0,r=0;
68                 getr(x,y,r);
69                 int cnt=0;
70                 for(int i=0;i<n;++i)
71                 {
72                     if(fabs(r*r- ((nu[i].x-x)*(nu[i].x-x)+(nu[i].y-y)*(nu[i].y-y)) )<=1e-6)
73                         ++cnt;
74                 }
75                 if(cnt*2>=n)
76                 {
77                     printf("%lf %lf %lf\n",x,y,r);
78                     break;
79                 }
80             }
81         }
82     }
83     return 0;
84 }