hdu1007——Quoit Design

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31780    Accepted Submission(s): 8352

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

 

Source

ZJCPC2004

 

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Statistic | Submit | Discuss | Note

求平面上的最近点对，分治大法好

简单介绍思路，先把所有点按x坐标从小到大排好序，然后取中间点分成2份，分别求出左右两份里的最小值，最近点对可以在左边，在右边，或者横跨两部分，前面两者很好求，关键是第三者，设由前两者求出的最小值为x，那么我们可以找出横坐标距mid点在x之内的点，然后求出最小距离，进一步优化，我们可以把这些进行y坐标排序，如果对于当前i和j两个点，他们纵坐标距离已经大于等于x了，那么显然不是最近距离了（最好情况是x坐标相同，但这时也没有缩短最小距离）

#include<map>
#include<set>
#include<list>
#include<stack>
#include<queue>
#include<vector>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

typedef struct
{
    double x, y;
}point;
point node[100100];
int pos[100100];

int cmpx(point a, point b)
{
    return a.x < b.x;
}

int cmpy(int a, int b)
{
    return node[a].y < node[b].y;
}

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

double min(double a, double b)
{
    return a < b ? a : b;
}

double min_distance(int low, int high)
{
    //2个点
    if(low == high - 1)
        return dist(node[low], node[high]);
    if(low == high - 2)
        return min(dist(node[low], node[low + 1]), min( dist(node[low], node[high]), dist(node[low + 1], node[high])) );
    int mid = (low + high) >> 1;
    double cur_closed = min( min_distance(low, mid), min_distance(mid + 1, high) );
    int cnt = 0;
    for(int i = low; i <= high; i++)
    {
        if(node[mid].x - cur_closed <= node[i].x && node[mid].x + cur_closed >= node[i].x)
        {
            pos[cnt++] = i;
        }
    }
    sort(pos, pos + cnt, cmpy);
    for(int i = 0; i < cnt; i++)
    {
        for(int j = i + 1; j < cnt; j++)
        {
            if(node[pos[j]].y - node[pos[i]].y >= cur_closed)
            {
                break;
            }
            cur_closed = min(cur_closed, dist(node[pos[j]], node[pos[i]]));
        }
    }
    return cur_closed;
}

int main()
{
    int n;
    while(~scanf("%d", &n), n)
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf", &node[i].x, &node[i].y);
        }
        sort(node, node + n, cmpx);
        double ans = min_distance(0, n - 1);
        printf("%.2lf\n", ans / 2);
    }
    return 0;
}