贪心——D

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



题目大意：
就是给你n组数据和圆的半径d，让你在x轴上画半径为d的圆，问：如果将所有的点都画进去，最少需要多少个圆，这个题目和导弹拦截有点像，不过更加简单

思路：
就是先判断d是不是大于等于0，如果d<0，肯定是输出-1的，
之后输入数字，如果有坐标的纵坐标比d还要大，那么也是不对的也要输出-1
之后对坐标进行处理，把每一个坐标在x轴上的范围标记出来，并进行排序，先排右边的位置，右边位置越小就排在越前面，因为我们是要从横坐标左边往右边排
如果右边相同，就排左边，左边大的先排，因为区间范围小的肯定可以把区间范围大的包括进去，反之则不行。
排完序之后
就开始画圈圈。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <iostream>
using namespace std;
const int maxn=1010;
struct node
{
    double l,r;
}exa[maxn];
bool cmp(node a,node b)
{
    if(a.r==b.r) return a.l>b.l;
    return a.r<b.r;
}

int main()
{
    int n,cnt=0;;
    double d,a,b;
    while(scanf("%d%lf",&n,&d)!=EOF&&(n+d))
    {
        bool flag=0;
        if(d>=0) flag=1;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&a,&b);
            if(b>d) flag=0;
            if(flag)
            {
                exa[i].l=a-sqrt(d*d-b*b);
                exa[i].r=a+sqrt(d*d-b*b);
            }
        }
        sort(exa,exa+n,cmp);
        int ans=-1;
        if(flag)
        {
            ans=1;
            double maxr=exa[0].r;
            for(int i=1;i<n;i++)
            {
                if(exa[i].l>maxr)
                {
                    ans++;
                    maxr=exa[i].r;
                }
            }
        }
        cout << "Case " << ++cnt << ": " << ans << endl;
    }
    return 0;
}