算法：
1 排序 算出每個(gè)島可安雷達(dá)的最左邊坐標(biāo)和最右邊坐標(biāo) 左座標(biāo)按升序排列 左相同時(shí)右按照降序排列
2 貪心選取 更新最右邊的坐標(biāo) 如果下一個(gè)的左座標(biāo)大于當(dāng)前最右邊 數(shù)目加一 否則的話更新最右邊的值

排序的時(shí)候因?yàn)槭荄OUBLE 所以那個(gè)qsort還要小小的注意下寫(xiě)法

代碼如下
/**********************
Author: WHU_Victordu
Created Time: 2007-12-26
File Name: pku1328.cpp
  Description: 
   **************************/
#include <stdio.h>
#include <stdlib.h>
#include <math.h>

typedef struct Point
{
        double left;
        double right;
}points;

int cmp(const void *p1, const void *p2)
{
   Point I1 = *(Point *)p1;
   Point I2 = *(Point *)p2;
   double temp = I1.left - I2.left;
   if (temp > 0.0) return 1;
   else if (temp < 0.0) return -1;
   else {
      temp = I2.right - I1.right;
      if (temp > 0.0) return 1;
      else return -1;
   }
}

int main()
{
    Point points[1005];

    int n,d,x[1005],y[1005],i,dd,cnt,flag;
    int t=0;
    double sqrtdy,right;
   
    while(scanf("%d%d",&n,&d)&&!(n==0&&d==0))
    {
      t++;
      flag=0;
      for(i=0;i<n;i++)
      {
       scanf("%d%d",&x[i],&y[i]);
       if(y[i]>d) flag=1;
       }
      if(flag)
      { 
         printf("Case %d: %d\n",t, -1);
         continue;
      }
     
      dd=d*d;
      for (i = 0; i < n; i++)
       {
         sqrtdy = sqrt(dd - y[i]*y[i]);
         points[i].left = x[i]-sqrtdy;
         points[i].right =x[i]+ sqrtdy;
      }

     
      qsort(points,n,sizeof(Point),cmp);
     
      for(i=0;i<n;i++)
      {
        printf("%d %d\n",points[i].left,points[i].right);
      }
     
      cnt=1;
      right=points[0].right;
      for(i=1;i<n;i++)
      {
        if(points[i].left>right)
        {
         cnt++;
         right=points[i].right;
        }
        else {
            if (points[i].right < right)
               right = points[i].right;
         }
      }
     
    printf("Case %d: %d\n",t,cnt);
     
    }
   
}