Radar Installation（贪心）

Radar Installation

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

代码如下：

#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{
    double l,r;
} a[1111];
bool cmp(node a,node b)
{
    return a.l<b.l;
}
int main()
{
    int n,i,j,flag,sum=0;
    double d,x,y;
    while(cin>>n>>d&&n&&d)
    {
        sum++;
        flag=1;
        for(i=0; i<n; i++)
        {
            cin>>x>>y;
            if(y>d)
            {
                flag=-1;
            }
            a[i].l=x-sqrt(d*d-y*y);
            a[i].r=x+sqrt(d*d-y*y);
        }
        if(flag==-1)
        {
            cout<<"Case "<<sum<<": "<<flag<<endl;
            continue;
        }
        sort(a,a+n,cmp);
        for(i=1;i<n;i++)
            if(a[i-1].r>=a[i].l)
        {
            a[i].l=max(a[i].l,a[i-1].l);
            a[i].r=min(a[i].r,a[i-1].r);
        }
        else  flag++;
        cout<<"Case "<<sum<<": "<<flag<<endl;
    }
    return 0;
}