Quoit Design - HDU 1007 分治

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本文解析了一个名为QuoitDesign的游戏算法题，该题目要求玩家找到能够恰好套住一个玩具的最大环的半径。文章提供了详细的输入输出样例及AC代码实现。

Quoit Design

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30473 Accepted Submission(s): 8008

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

Sample Output

0.71
0.00
0.75

思路：新手上路，鉴于我的理解不够深刻就不误导大家了，就只提供下代码吧。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct node
{ double x,y;
}dian[100010];
int a[100010];
bool cmpx(node a,node b)
{ return a.x<b.x;}
bool cmpy(int a,int b)
{ return dian[a].y<dian[b].y;}
double dis(int i,int j)
{ return sqrt((dian[i].x-dian[j].x)*(dian[i].x-dian[j].x)+(dian[i].y-dian[j].y)*(dian[i].y-dian[j].y));}
double fen(int l,int r)
{ int i,j,num=0;
  if(l+1==r)
   return dis(l,r);
  if(l+2==r)
   return min(dis(l,r),min(dis(l,l+1),dis(l+1,r)));
  int mi=(l+r)/2;
  double ans=min(fen(l,mi),fen(mi+1,r));
  for(i=l;i<=r;i++)
    if(fabs(dian[i].x-dian[mi].x)<=ans)
     a[++num]=i;
  sort(a+1,a+1+num,cmpy);
  for(i=1;i<=num;i++)
   for(j=i+1;j<=num;j++)
   { if(dian[a[j]].y-dian[a[i]].y>=ans)
      break;
     ans=min(ans,dis(a[i],a[j]));
   }
  return ans;
}
int main()
{ int n,i,j,k;
  while(~scanf("%d",&n) && n)
  { for(i=1;i<=n;i++)
     scanf("%lf%lf",&dian[i].x,&dian[i].y);
    sort(dian+1,dian+1+n,cmpx);
    printf("%.2f\n",fen(1,n)/2);
  }
}