[HDU1007]Quoit Design

题目描述 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.

输入描述 Input Description

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.

输出描述 Output Description

For each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places. 

样例输入 Sample Input

2

0 0

1 1

2 1

1 1

1 3

-1.5 0

0 0

0 1.5

0

样例输出 Sample Output

0.71

0.00

0.75

数据范围及提示 Data Size & Hint

之前的一些废话：是时候准备会考了。。

题解：求最近点对。首先把平面划分成两个部分，递归求出两个部分的答案为ans，然后，把离分割线距离小于ans/2的点一左一右算距离更新ans.

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
typedef long long LL;
#define mem(a,b) memset(a,b,sizeof(a))
typedef pair<int,int> PII;
const int maxn=100010;
const double oo=2147483647;
struct Point
{
    double x,y;
    Point() {}
    Point(double _1,double _2):x(_1),y(_2){}
    bool operator < (const Point &s)const
    {
        if(x==s.x)return y<s.y;
        return x<s.x;
    }
}p[maxn];
int n;double x,y;
double dis(double a,double b,double c,double d){
    return sqrt((c-a)*(c-a)+(d-b)*(d-b));
}
double mdis(int l,int r)
{
    if(l==r)return 0;
    if(l+1==r)return dis(p[l].x,p[l].y,p[r].x,p[r].y);
    double ret=oo;
    int mid=(l+r)>>1;
    ret=min(mdis(l,mid),mdis(mid,r));
    for(int i=mid-1;i>=l && p[mid].x-p[i].x<ret;i--)
        for(int j=mid+1;j<=r && p[j].x-p[mid].x<ret && fabs(p[i].y-p[j].y)<ret;j++)
            ret=min(ret,dis(p[i].x,p[i].y,p[j].x,p[j].y));
    return ret;
}
int main()
{
    while(scanf("%d",&n)!=EOF && n)
    {
        for(int i=0;i<n;i++)scanf("%lf%lf",&x,&y),p[i]=Point(x,y);
        sort(p,p+n);
        printf("%.2lf\n",mdis(0,n-1)/2);
        for(int i=0;i<n;i++)p[i]=Point(0,0);
    }
    return 0;
}

总结：