Hdu 1007 Quoit Design

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 64492    Accepted Submission(s): 17082

 

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

大意就是找兩個最近的點的距離的一半，不過我不會誒，分治思想，把各個點從中間分開，找到兩段個點離中間值的最小值，然後合併他們，當然有可能出現最短點在同一邊的可能，這時候你需要更新最短的距離。

在搜邊的時候，如果這條邊大於中間的邊+當前最小值或者小於中間的邊-當前最小值的話，不予討論絕對不是最小的，

然後按y排序，更新最小值，如果大於ansbreak；

#include <iostream>
#include<cstdio>
#include<string.h>
#include<math.h>
#include<algorithm>
#define ll long long
const int mod=7;
using namespace std;
struct node
{
    double x;
    double y;
}ac[100005];
int q[100005];
double dist(node a,node b)
{
    return sqrt((a.y-b.y)*(a.y-b.y)+(a.x-b.x)*(a.x-b.x));
}
bool cmpx(node a,node b)
{
    return a.x<b.x;
}
bool cmpy(int a,int b)
{
    return  ac[a].y<ac[b].y;
}
double find(int l,int r)
{
    if(l+1==r)
        return dist(ac[l],ac[r]);
        if(l+2==r)
            return min(dist(ac[l],ac[r]),min(dist(ac[l],ac[l+1]),dist(ac[l+1],ac[r])));
    int mid=(l+r)>>1;
    double  ans=min(find(l,mid),find(mid+1,r));
    int i,j,cnt=0;
    for(int i=l;i<=r;i++)
    {
        if(ac[i].x>=ac[mid].x-ans&&ac[i].x<=ac[mid].x+ans)
        {
            q[cnt++]=i;
        }
    }
    sort(q,q+cnt,cmpy);
    for(int i=0;i<cnt;i++)
    {
        for(int j=i+1;j<cnt;j++)
        {
            if(ac[q[j]].y-ac[q[i]].y>ans)break;
            ans=min(ans,dist(ac[q[j]],ac[q[i]]));
        }
    }
    return ans;
}
int main()
{
     ll t;
     while(scanf("%lld",&t)&&t!=0)
     {
         memset(q,0,sizeof(q));
         for(int i=0;i<t;i++)
         {
             scanf("%lf%lf",&ac[i].x,&ac[i].y);
         }
         sort(ac,ac+t,cmpx);
         printf("%.2lf\n",find(0,t-1)/2);
     }
    return 0;
}