ACM-计算几何之Quoit Design——hdu1007 zoj2107

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 28539    Accepted Submission(s): 7469

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

 

Author

CHEN, Yue

 

题目：

pid=1007">hdu 1007     。      zoj  2107

这道题。类型：求近期点对。

求平面 近期点对的方法。就是分治法。

先将点分成两个区间，如果S1。S2，然后分别求S1内近期点对的点d1，S2内近期点对的点d2

再求S1与S2内近期点对 d=min（d1,d2）

可是。不能忘记，近期点对可能是 一个点在S1一个点在S2。

接下来就是比較精华的部分：

所求的点的位置，一定在于  mid-d,mid+d 之间。

然后。就在这个区间開始找点，并不断更新d值，最后就能够得到d了。

这道题，终于要求半径，所以还要除以2。

/**************************************
***************************************
*        Author:Tree                  *
*From :http://blog.csdn.net/lttree    *
* Title : Quoit Design                *
*Source: hdu 1007 zoj 2107            *
* Hint  : 计算几何——近期点对         *
***************************************
**************************************/
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define N 100001
struct Point
{
    double x,y;
}p[N];
int arr[N];
double Min(double a,double b)
{
    return a<b?a:b;
}
// 求两点之间的距离
double dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
// 依据点横坐标or纵坐标排序
bool cmp_y( int a,int b)
{
    return p[a].y<p[b].y;
}
bool cmp_x( Point a,Point b)
{
    return a.x<b.x;
}
// 求近期点对
double close_pair( int l,int r )
{
    // 推断两个点和三个点的情况
    if( r==l+1 )    return dis( p[l],p[r] );
    else if( r==l+2 )   return Min( dis(p[l],p[r]),Min( dis(p[l],p[l+1]),dis(p[l+1],p[r]) ) );

    int mid=(l+r)>>1;
    double ans=Min(close_pair(l,mid),close_pair(mid+1,r));

    int i,j,cnt=0;
    // 假设 当前p[i]点 横坐标位于 范围（中点横坐标-ans，中点横坐标+ans）位置内，则记录点的序号
    for(i=l; i<=r; ++i)
        if(  p[i].x>=p[mid].x-ans && p[i].x<=p[mid].x+ans  )
            arr[cnt++]=i;
    // 依照纵坐标由小到大 对于arr数组内点进行排序
    sort(arr,arr+cnt,cmp_y);
    for(i=0; i<cnt; i++)
        for(j=i+1; j<cnt; j++)
        {
            if(p[arr[j]].y-p[arr[i]].y>=ans) break;
            ans=Min(ans,dis(p[arr[i]],p[arr[j]]));
        }

    return ans;
}

int main()
{
    int i,n;
    while( scanf("%d",&n)!=EOF && n)
    {
        for(i=0;i<n;++i)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        // 先将全部点依照横坐标由小到大排序
        sort(p,p+n,cmp_x);
        printf("%.2lf\n",close_pair(0,n-1)/2.0);
    }
    return 0;
}