zoj 2107 Quoit Design(최근점 쌍)

Quoit Design
Time Limit: 5 Seconds      
Memory Limit: 32768 KB
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: 
Zhejiang Provincial Programming Contest 2004
제목:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2107
제목: n점을 줄게, 가장 가까운 두 점의 거리의 반을 빌어...
분석: 최근점 맞다, 처음 들어보는 것 같다==, 분치로 할 수 있다. 먼저 x축에 따라 정렬한 다음에 매번 평면을 반으로 나누어 각각 해답을 구한다. 한 번의 합병에 대해 먼저 구분선 거리가 현재의 최소치보다 작은 점을 찾아낸 다음에 이 점들을 y축에 따라 순서를 정한다. 사실 x축도 마찬가지다. 폭력적으로 이 점들을 구하는 가장 가까운 점은 매번 y좌표의 차이가 답보다 크다. 직접break.내가 쓴 이것은 상당히 폭력적이지만, 그래도 A다, 900++ms
시간이 나면 다시 최적화하세요.
코드:

#include<cmath>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int mm=111111;
typedef double diy;
struct point
{
    diy x,y;
    point(){}
    point(diy _x,diy _y):x(_x),y(_y){}
}g[mm];
int q[mm];
int n;
bool cmp(point p,point q)
{
    return p.x<q.x;
}
bool cmpy(int a,int b)
{
    return g[a].y<g[b].y;
}
diy Dis(point P,point Q)
{
    return sqrt((P.x-Q.x)*(P.x-Q.x)+(P.y-Q.y)*(P.y-Q.y));
}
diy NearPoint(int l,int r)
{
    if(l>=r)return 1e30;
    int i,j,n=0,m=(l+r)>>1;
    diy a=min(NearPoint(l,m),NearPoint(m+1,r));
    for(i=m;i>=l;--i)
        if(g[m].x-g[i].x<a)q[n++]=i;
        else break;
    for(i=m+1;i<=r;++i)
        if(g[i].x-g[m].x<a)q[n++]=i;
        else break;
    sort(q,q+n,cmpy);
    for(i=0;i<n;++i)
        for(j=i+1;j<n;++j)
            if(g[q[j]].y-g[q[i]].y<a)a=min(a,Dis(g[q[i]],g[q[j]]));
            else break;
    return a;
}
int main()
{
    while(scanf("%d",&n),n)
    {
        for(int i=0;i<n;++i)
            scanf("%lf%lf",&g[i].x,&g[i].y);
        sort(g,g+n,cmp);
        printf("%.2lf
",NearPoint(0,n-1)/2);
    }
    return 0;
}