hdu 1007 Quoit Design (最近点对)

http://acm.hdu.edu.cn/showproblem.php?pid=1007

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43467    Accepted Submission(s): 11301

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
   
   
   
   

 

题意：给定平面上n个点，找其中的一对点，使得在n个点的所有点对中，该点对的距离最小。

最近点对算法题，详情请点击

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>

using namespace std;

#define N 110000
#define INF 0x3f3f3f3f
#define met(a, b) memset (a, b, sizeof(a))

struct node
{
    double x, y;
}stu1[N], stu2[N];

bool cmpx (node a, node b)
{
    return a.x < b.x;
}

bool cmpy (node a, node b)
{
    return a.y < b.y;
}

int n;

double dist (node a, node b)
{
    return sqrt ((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}

double mindist (int l, int r)
{
    if (l+1 == r) return dist(stu1[l], stu1[r]);///剩两个点的情况
    if (l+2 == r)///剩三个点的情况
        return min (dist(stu1[l], stu1[l+1]), min (dist(stu1[l], stu1[r]), dist(stu1[l+1], stu1[r])));

    int mid = (l + r) / 2;

    double d = min (mindist (l, mid), mindist (mid+1, r));///左右递归

    int k = 0;
    for (int i=l; i<=r; i++)
        if (fabs (stu1[i].x-stu1[mid].x) < d)
            stu2[k++] = stu1[i];///找到离中间线距离小于两遍的最短距离的点

    sort (stu2, stu2+k, cmpy);///y坐标排序

    for (int i=0; i<k; i++)
    for (int j=i+1; j<k; j++)
    {///两边各有一点之间的距离小于两遍的最小距离
        if (dist(stu2[i], stu2[j]) >= d)
            break;
        d = min (d, dist(stu2[i], stu2[j]));
    }
    return d;
}

int main ()
{
    while (scanf ("%d", &n), n)
    {
        for (int i=0; i<n; i++)
            scanf ("%lf%lf", &stu1[i].x, &stu1[i].y);

        sort (stu1, stu1+n, cmpx);

        printf ("%.2f\n", mindist (0, n-1)/2);
    }
    return 0;
}