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
1 #include <cstdio>
2 #include <cmath>
3 #include <cstring>
4 #include <cstdlib>
5 #include <iostream>
6 #include <sstream>
7 #include <algorithm>
8 #include <string>
9 #include <queue>
10 #include <vector>
11 using namespace std;
12 const int maxn= 1e5+10;
13 const double eps= 1e-6;
14 const int inf = 0x3f3f3f3f;
15 typedef long long ll;
16 struct point
17 {
18 double x,y;
19 }a[maxn];
20 int n;
21 double dis(point a,point b) //两点间距离
22 {
23 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
24 }
25 bool waijie(point p1,point p2,point p3,point &ans) //引用修改圆心的值
26 {
27 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
28 double Bx = p2.x - p1.x, By = p2.y - p1.y; //外接圆板子
29 double Cx = p3.x - p1.x, Cy = p3.y - p1.y;
30 double D = 2 * (Bx * Cy - By * Cx);
31 double cx = (Cy * (Bx * Bx + By * By) - By * (Cx * Cx + Cy * Cy)) / D + p1.x;
32 double cy = (Bx * (Cx * Cx + Cy * Cy) - Cx * (Bx * Bx + By * By)) / D + p1.y;
33 ans.x=cx,ans.y=cy;
34 return true;
35 }
36 bool check(point mid,double d) //检查是否有n/2个点在外接圆上
37 {
38 int ans=0;
39 for(int i=1;i<=n;i++)
40 {
41 if(fabs(dis(a[i],mid)-d)<=eps)
42 ans++;
43 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
44 return false;
45 }
46 if(ans*2>=n)
47 return true;
48 return false;
49 }
50 int main()
51 {
52 int t;
53 scanf("%d",&t);
54 while(t--)
55 {
56 scanf("%d",&n);
57 for(int i=1;i<=n;i++)
58 scanf("%lf%lf",&a[i].x,&a[i].y);
59 if(n<=2) //n小于等于4特判
60 {
61 printf("%lf %lf %lf\n",a[1].x,a[1].y,0.0);
62 continue;
63 }
64 else if(n<=4)
65 {
66 printf("%lf %lf %lf\n",(a[1].x+a[2].x)/2,(a[1].y+a[2].y)/2,dis(a[1],a[2])/2);
67 continue;
68 }
69 while(true)
70 {
71 point aa=a[rand()%n+1],bb=a[rand()%n+1],cc=a[rand()%n+1]; //随机产生3个点
72 point xin;
73 if(!waijie(aa,bb,cc,xin))
74 continue;
75 double r=dis(aa,xin);
76 if(check(xin,r))
77 {
78 // printf("%lf %lf\n",aa.x,aa.y);
79 // printf("%lf %lf\n",bb.x,bb.y);
80 // printf("%lf %lf\n",cc.x,cc.y);
81 printf("%lf %lf %lf\n",xin.x,xin.y,r);
82 break;
83 }
84 }
85 }
86 return 0;
87 }