hdu 1007 Quoit Design

最新推荐文章于 2024-07-18 12:00:00 发布

转载最新推荐文章于 2024-07-18 12:00:00 发布 · 101 阅读

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原文链接：http://www.cnblogs.com/52why/p/7478098.html

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#java

本文介绍了一个通过计算几何方法解决的最大圆环半径问题。该问题要求在一个平面上找到能够恰好包围一个玩具点的最大半径圆环，且不能同时包围两个或以上的玩具点。文章提供了完整的C++代码实现，并通过输入多个测试用例验证了算法的有效性。

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 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 using namespace std;
 7 #define INF 0x7f7f7f
 8 int n;
 9 struct Point
10 {
11     double x,y;
12 }p[100001];
13 bool cmp(Point a,Point b)
14 {
15     if(a.x==b.x)
16     {
17         return a.y<b.y;
18     }
19     return a.x<b.x;
20 }
21 double dis(Point a,Point b)
22 {
23     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
24 }
25 double getMin(int i)
26 {
27      double d=999999,di=0;
28      for(int j=i+1;j<n;j++)
29      {
30          di=dis(p[i],p[j]);
31          if(d>di)d=di;
32          else break;
33      }
34      return d;
35 }
36 int main()
37 {
38     int i,j,k;
39     double ans,f;
40     while(scanf("%d",&n)!=EOF)
41     {
42         if(!n)break;
43         for(i=0;i<n;i++)
44         {
45             scanf("%lf%lf",&p[i].x,&p[i].y);
46         }
47         sort(p,p+n,cmp);
48         ans=INF;
49         for(i=0;i<n-1;i++)
50         {
51             f=getMin(i);
52             if(ans>f)ans=f;
53         }
54         printf("%.2lf\n",ans/2.0);
55     }
56     return 0;
57 }

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55105 Accepted Submission(s): 14543

Problem Description

Have you ever played quoit in a playground? Quoit is a game in which flat rings are pitched at some toys, with all the toys encircled awarded.
In the field of Cyberground, the position of each toy is fixed, and the ring is carefully designed so it can only encircle one toy at a time. On the other hand, to make the game look more attractive, the ring is designed to have the largest radius. Given a configuration of the field, you are supposed to find the radius of such a ring.

Assume that all the toys are points on a plane. A point is encircled by the ring if the distance between the point and the center of the ring is strictly less than the radius of the ring. If two toys are placed at the same point, the radius of the ring is considered to be 0.

Input

The input consists of several test cases. For each case, the first line contains an integer N (2 <= N <= 100,000), the total number of toys in the field. Then N lines follow, each contains a pair of (x, y) which are the coordinates of a toy. The input is terminated by N = 0.

Output

For each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places.

Sample Input

2 0 0 1 1 2 1 1 1 1 3 -1.5 0 0 0 0 1.5 0

Sample Output

0.71 0.00 0.75

转载于:https://www.cnblogs.com/52why/p/7478098.html