最近对问题,分治法求解
来源:互联网 发布:西游之路坐骑进阶数据 编辑:程序博客网 时间:2024/05/17 03:24
最近对问题,即从一堆点中,找出离得最近的两个点之间的距离
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.
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
20 01 121 11 13-1.5 00 00 1.50
<span style="font-family:Courier New;">#include <cstdio>#include <cmath>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn=100000+5;const int INF=1<<31-1;struct Point{ double x,y;};Point p[maxn];Point pt[maxn];bool cmp(Point &p1,Point &p2){ return p1.x<p2.x;}bool cmp2(Point &p1,Point &p2){ return p1.y<p2.y;}double getDis(Point &p1,Point &p2){ return sqrt( pow(p1.x-p2.x,2) + pow(p1.y-p2.y,2) );}double getClosest(int left,int right){ if(left==right) return INF; if(right-left==1) return getDis(p[left],p[right]); int mid=(left+right)/2; double d1=getClosest(left,mid); double d2=getClosest(mid+1,right); double d=(d1<d2?d1:d2); int index=0; for(int i=mid;i>=left;i--) { if(p[mid].x-p[i].x>d) break; else pt[index++]=p[i]; } for(int i=mid+1;i<=right;i++) { if(p[i].x-p[mid].x>d) break; else pt[index++]=p[i]; } sort(pt,pt+index,cmp2); //这里一定要排序,排序后,按照鸽笼原理,dis最多更新8次,可减少计算 for(int i=0;i<index-1;i++) { for(int j=i+1;j<index;j++) { if(pt[j].y-pt[i].y>d) break; else { double dis=getDis(pt[i],pt[j]); if(dis<d) d=dis; } } } return d;}int main(){ int n; while(scanf("%d",&n)!=EOF && n) { for(int i=0;i<n;i++) { scanf("%lf%lf",&p[i].x,&p[i].y); } sort(p,p+n,cmp); double ans=getClosest(0,n-1); printf("%.2lf\n",ans/2); } return 0;}</span>
0 0
- 最近对问题,分治法求解
- 分治法求解最近点对问题
- 分治法求解最近点对
- 分治法-最近对问题
- hdu 1007 经典最近点对 (分治法求解)
- 用分治法求解三维空间中的最近点对
- 分治法求最近点对问题
- 分治法:平面最近点对问题
- 分治法最近点对问题
- 分治法求最近点对问题
- 最近对问题(分治法)
- 分治法实现最近对问题
- 分治法求解最近点距
- POJ 3714 最近点对问题 分治法
- 用分治法解决最近点对问题
- 分治策略 求 最近点对问题
- 最近点对问题(分治)
- zoj2107 最近点对问题 分治思想
- HDOJ 5651-xiaoxin juju needs help【字符的全排列】
- 红黑树代码实践
- 归园田居 东晋 陶渊明 注音版
- AOJ2249Contest Page(最短路径spfa)
- 借助SystemBarTint实现透明标题栏(沉浸式)
- 最近对问题,分治法求解
- Android Studio 2.1 Preview 4 发布,以及新增的两点攺动.
- adb相关问题
- 合并模拟器和真机调试的静态库
- 堆排序
- UVA 11090 图论加二分
- ubuntu tftp服务搭建
- CentOS6 Shell脚本/bin/bash^M: bad interpreter错误解决方法
- IOS百度地图自定义大头针和气泡