poj 1328 Radar Installation(贪心+快排)

Description

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates. 
 
Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases.

The input is terminated by a line containing pair of zeros

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Sample Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

题意：将一条海岸钱看为X轴，X轴的上方为大海，海上有许多岛屿，给出岛屿的位置与雷达的覆盖半径，要求在海岸线上建雷达，
　　  在雷达能够覆盖所有岛的基础上，求最少需要多少雷达。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <cstring>
 #include <cstdlib>
 using namespace std;
 #define MAX 1005
 struct sea
 {
     double left;
     double right;
 } a[];
 bool operator < (sea A,sea B)
 {
     return A.left<B.left;
 }
 int main()
 {
     int n,k=;
     double d;
     while(cin>>n>>d&&(n||d))
     {
         bool flag=false;
         for(int i=; i<n; i++)
         {
             double x,y;
             cin>>x>>y;
             if(fabs(y)>d)
                 flag=true;
             else
             {  //计算区间
                 a[i].left=x*1.0-sqrt(d*d-y*y);
                 a[i].right=x*1.0+sqrt(d*d-y*y);
             }
         }
         printf("Case %d: ",k++);
         if(flag)
             printf("-1\n");
         else
         {
             int ans=; //雷达初始化
             sort(a,a+n); //  排序
             double s=a[].right;
             for(int i=; i<n; i++)
             {
                 if(a[i].left>s)
                 {
                     ans++;    //雷达加一
                     s=a[i].right; // 更新右端点
                 }
                 else if(a[i].right<s)
                     s=a[i].right;
             }
             printf("%d\n",ans);
         }
     }
     return ;
 }