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

題意：

??假設海岸線是一條無限延伸的直線。陸地在海岸線的一側，而海洋在另一側。每一個小的島嶼是海洋上的一個點。雷達坐落于海岸線上，只能覆蓋d距離，所以如果小島能夠被覆蓋到的話，它們之間的距離最多為d。題目要求計算出能夠覆蓋給出的所有島嶼的最少雷達數目。

解題思路：

??我們假設島嶼i它的x坐標為island[i][0]，而y坐標為island[i][1]，那么有以下幾種情況是invalide的，即輸出-1的情況：
1.island[i][1]<0
2.abs(island[i][1])>d
3.d<0
其他的情況，應該就是正常情況，進入計算最小雷達數目。

image.png

如上圖，紅色的點為島嶼，那么能夠覆蓋到此島嶼的雷達所在的區間，應該就是以該島嶼為圓心的圓與x軸交點所在的區間。

這樣，我們就可以計算出所有島嶼的雷達所在的區間，得到一個區間數組。

我們將這個數組按照區間左部分進行排序，那么重疊部分就表明這些島嶼的雷達可以共用一個。從而計算出最終解。

可以把所有的島的坐標通過x(+ -)sqrt(d*d-y*y)得出可以掃射到它的雷達的區域范圍，在對右坐標進行排序。

但是千萬不要忘記還有右坐標相等的情況，這種時候你知道肯定要選區間小的范圍（一定可以掃射到區間范圍大的島），即讓左坐標大的排在前面。

拍完序后，你就要用到貪心法，每次都貪心的選擇最右端的點作為判斷點，循環到最后，可以得出最少的雷達數。

#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)      //結構體排序
{
    if(a.right==b.right) return a.left>b.left; 
        else
    return a.right<b.right;
}
int acount,n; 
void search(double maxn)      //找到最少的雷達數
{
    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)                      //判斷是否可以掃射到島
                o++;
            b[i].left=x-sqrt((d*d)-(y*y));         //預處理，變成區間
            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;
}