Hdu1007 Quoit Design

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58163    Accepted Submission(s): 15416

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

Recommend

JGShining

题意：求平面上两个点的最短距离/2

分析：分治经典题，注意边界的处理.

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

using namespace std;

int n;

struct node
{
    double x,y;
} e[100010],p[100010];

bool cmp1(node a,node b)
{
    return a.x < b.x;
}

bool cmp2(node a,node b)
{
    return a.y < b.y;
}

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 solve(int l,int r)
{
    if (l >= r)
        return 1e20;
    if (r == l + 1)
        return dist(e[l],e[r]);
    if (r == l + 2)
        return min(dist(e[l],e[l + 1]),min(dist(e[l + 1],e[r]),dist(e[l],e[r])));
    int mid = (l + r) >> 1,cnt = 0;
    double res = min(solve(l,mid - 1),solve(mid + 1,r));
    for (int i = l; i <= r; i++)
        if (fabs(e[mid].x - e[i].x) <= res)
            p[++cnt] = e[i];
    sort(p + 1,p + 1 + cnt,cmp2);
    for (int i = 1; i <= cnt; i++)
        for (int j = i + 1; j <= cnt; j++)
        {
            if (p[j].y - p[i].y > res)
                break;
            res = min(res,dist(p[j],p[i]));
        }
    return res;
}

int main()
{
    while (scanf("%d",&n) != EOF && n)
    {
        for (int i = 1; i <= n; i++)
            scanf("%lf%lf",&e[i].x,&e[i].y);
        sort(e + 1,e + 1 + n,cmp1);
        printf("%.2lf\n",solve(1,n)/2.0);
    }

    return 0;
}