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

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

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.

3 2

1 2

-3 1

2 1

1 2

0 2

0 0

Case 1: 2

Case 2: 1

该题题意是为了求出能够覆盖所有岛屿的最小雷达数目

这题一开始根本没有贪心思路，想不到

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<queue>

#include<cctype>

using namespace std;

struct node

{

    double left,right;

}qu[];

int cmp(node a,node b)

{

    return a.right<b.right;

}

int main() {

    int n,k=;

    double r;

    while(scanf("%d%lf",&n,&r)!=EOF){

        if (n== && r== ) break;

        int flag=;

        double x,y;

        for (int i= ;i<n ;i++){

            scanf("%lf%lf",&x,&y);

            if (!flag) continue;

            if (y>r) {

                flag=;

                continue;

            }

            qu[i].left=x-sqrt(r*r-y*y);

            qu[i].right=x+sqrt(r*r-y*y);

        }

        sort(qu,qu+n,cmp);

        printf("Case %d: ",k++);

        if (!flag) {

            printf("-1\n");

            continue;

        }

        int sum=;

        double temp=-;

        for (int i= ; i<n ;i++){

            if (qu[i].left>temp){

                sum++;

                temp=qu[i].right;

            }

        }

        printf("%d\n",sum);

    }

    return ;

}