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hdu 1007 Quoit Design 分治求最近点对

2024-07-07 01:22:43   227人阅读

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29344    Accepted Submission(s): 7688


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
题目大意：给n个点，求最近点对距离的一半。
按x值排序，分治法二分递归搜索，合并的时候注意一下把那些fabs(p[i].x-p[mid].x)<=ans的点找出来，这些点中可能有更小的ans,把他们按y值排序,暴力两层循环更新ans,当p[j].y-p[i].y>=ans时没必要继续了。
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int maxn=100005;
 8 int N,f[maxn];
 9 struct Point
10 {
11     double x,y;
12 }p[maxn];
13 
14 double min(double a,double b){ return a<b?a:b;}
15 bool cmpx(const Point &a,const Point &b)
16 {
17     return a.x<b.x;
18 }
19 bool cmpy(const int a,const int b)
20 {
21     return p[a].y<p[b].y;
22 }
23 double dist(int a,int b)
24 {
25     return sqrt((p[a].x-p[b].x)*(p[a].x-p[b].x)+(p[a].y-p[b].y)*(p[a].y-p[b].y));
26 }
27 double BinarySearch(int l,int r)
28 {
29     if(l+1==r)
30     {
31         return dist(l,r);
32     }
33     if(l+2==r)
34     {
35         return min(min(dist(l,l+1),dist(l+1,l+2)),dist(l,l+2));
36     }
37     int mid=(l+r)>>1;
38     double ans=min(BinarySearch(l,mid),BinarySearch(mid+1,r));
39     int i,j,cnt=0;
40     for(i=l;i<=r;i++)
41     {
42         if(fabs(p[i].x-p[mid].x)<=ans)
43             f[cnt++]=i;
44     }
45     sort(f,f+cnt,cmpy);
46     for(i=0;i<cnt;i++)
47     {
48         for(j=i+1;j<cnt;j++)
49         {
50             if(p[f[j]].y-p[f[i]].y>=ans) break;
51             ans=min(ans,dist(f[i],f[j]));
52         }
53     }
54     return ans;
55 }
56 int main()
57 {
58     while(scanf("%d",&N),N)
59     {
60         for(int i=0;i<N;i++)
61             scanf("%lf %lf",&p[i].x,&p[i].y);
62         sort(p,p+N,cmpx);
63         printf("%.2lf\n",BinarySearch(0,N-1)/2);
64     }
65     return 0;
66 }

 

 


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