HDU 1007 Quoit Design 最近点对问题

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Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24618    Accepted Submission(s): 6542

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

20 01 121 11 13-1.5 00 00 1.50

 

Sample Output

0.710.000.75

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

Recommend

JGShining

给你n个点的坐标，让你求最近的两个点之间的距离。

如果枚举，肯定超时，稍加优化，便可AC。

#include<stdio.h>#include<algorithm>#include<math.h>#define inf 0x3f3f3fusing namespace std;struct Point{    double x,y;}p[100007];int cmp(const Point &a,const Point &b){    return a.x==b.x?a.x<b.x:a.y<b.y;}double dis(Point a,Point b){    return sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y));}int main(){    int n;    while(scanf("%d",&n),n)    {        for(int i=0;i<n;i++)            scanf("%lf%lf",&p[i].x,&p[i].y);        sort(p,p+n,cmp);        double ans=inf;        for(int i=0;i<n-1;i++)        {            double flag=inf,l;            for(int j=i+1;j<n;j++)            {                l=dis(p[i],p[j]);                if(flag>l)flag=l;//判断条件                else break;            }            if(ans>flag)                ans=flag;        }        printf("%.2f\n",ans/2);    }    return 0;}