【HDU 1007】Quoit Design（分治法）

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52560    Accepted Submission(s): 13857

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

 

思路：暴力循环所有情况，显然会超时，考虑分治的思想将问题分为子问题分别解决后合并。

代码：

#include<stdio.h>#include<math.h>#include<algorithm> using namespace std;struct point {double x;double y;}ar[100005],br[100005]; //br用于存放特殊情况的可疑点 double val,ans;bool cmp_x(struct point a,struct point b);//自定义比较函数（用在sort里） bool cmp_y(struct point a,struct point b);double dist(struct point a,struct point b);//计算距离 double solve(int l,int r); //解答函数，分治思想 int main(){int n,i,j;while(scanf("%d",&n)!=EOF&&n){for(i=0;i<n;i++) scanf("%lf %lf",&ar[i].x,&ar[i].y);sort(ar,ar+n,cmp_x);ans=solve(0,n-1);printf("%.2f\n",ans/2);}return 0;}bool cmp_x(struct point a,struct point b){if(a.x==b.x) return a.y<b.y;else return a.x<b.x;}bool cmp_y(struct point a,struct point b){if(a.y==b.y) return a.x<b.x;else return a.y<b.y;}double dist(struct point a,struct point 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(r-1==l) return dist(ar[l],ar[r]);  //只有两个点的情况 if(r-2==l) {                          //三个点的情况 double s1=dist(ar[l],ar[l+1]);double s2=dist(ar[l],ar[r]);double s3=dist(ar[l+1],ar[r]);return min(s1,min(s2,s3));}int mid=(l+r)/2;                      //分而治之  val=min(solve(l,mid),solve(mid+1,r)); //取两边距离的最小值 //特殊情况，若最近距离分别在分断处左右 int k=0,i,j; //筛选                          for(i=l;i<=r;i++){if(fabs(ar[i].x-ar[mid].x)<val){  //若x1-x2>=val（距离），直接跳过，反之存进br数组 br[k].x=ar[i].x;br[k].y=ar[i].y;k++;}}sort(br,br+k,cmp_y);  //按y从小到大排序 for(i=0;i<k-1;i++){for(j=i+1;j<k&&((br[j].y-br[i].y)<val);j++){//y已按从小到大排序，若(b[j].y-b[i].y)>=val，则后面的肯定更大。   val=min(val,dist(br[j],br[i]));}}return val;}