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

1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

题意：

  假设海岸线是一条无限延伸的直线。陆地在海岸线的一侧，而海洋在另一侧。每一个小的岛屿是海洋上的一个点。雷达坐落于海岸线上，只能覆盖d距离，所以如果小岛能够被覆盖到的话，它们之间的距离最多为d。题目要求计算出能够覆盖给出的所有岛屿的最少雷达数目。

解题思路：

  我们假设岛屿i它的x坐标为island[i][0]，而y坐标为island[i][1]，那么有以下几种情况是invalide的，即输出-1的情况：
1.island[i][1]<0
2.abs(island[i][1])>d
3.d<0
其他的情况，应该就是正常情况，进入计算最小雷达数目。

image.png

如上图，红色的点为岛屿，那么能够覆盖到此岛屿的雷达所在的区间，应该就是以该岛屿为圆心的圆与x轴交点所在的区间。

这样，我们就可以计算出所有岛屿的雷达所在的区间，得到一个区间数组。

我们将这个数组按照区间左部分进行排序，那么重叠部分就表明这些岛屿的雷达可以共用一个。从而计算出最终解。

可以把所有的岛的坐标通过x(+ -)sqrt(d*d-y*y)得出可以扫射到它的雷达的区域范围，在对右坐标进行排序。

但是千万不要忘记还有右坐标相等的情况，这种时候你知道肯定要选区间小的范围（一定可以扫射到区间范围大的岛），即让左坐标大的排在前面。

拍完序后，你就要用到贪心法，每次都贪心的选择最右端的点作为判断点，循环到最后，可以得出最少的雷达数。

#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#define MAX 1005
using namespace std;
struct note{
    double left,right;
}b[MAX];
bool cmp(note a,note b)      //结构体排序
{
    if(a.right==b.right) return a.left>b.left; 
        else
    return a.right<b.right;
}
int acount,n; 
void search(double maxn)      //找到最少的雷达数
{
    int i;
    for(i=1;i<n;i++)
        if(b[i].left>maxn)
        {
            maxn=b[i].right;        //都取最右端
            acount++;
        }
}
int main()
{
    double d,x,y;
    scanf("%d%lf",&n,&d);
    int t=0;
    while(n||d)
    {
        t++;
        int o=0;
        acount=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&x,&y);
            if(y>d)                      //判断是否可以扫射到岛
                o++;
            b[i].left=x-sqrt((d*d)-(y*y));         //预处理，变成区间
            b[i].right=x+sqrt((d*d)-(y*y));
        }
        if(!o)
        {
            sort(b,b+n,cmp);
            search(b[0].right);
            printf("Case %d: %d\n",t,acount+1);
        }
        else
            printf("Case %d: -1\n",t);
        scanf("%d%lf",&n,&d);
    }
    return 0;
}