ACM雷达安放问题

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<tex2html_verbatim_mark> distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d<tex2html_verbatim_mark> .

We use Cartesian coordinate system, defining the coasting is the x<tex2html_verbatim_mark> -axis. The sea side is above x<tex2html_verbatim_mark> -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<tex2html_verbatim_mark> - y<tex2html_verbatim_mark>coordinates.

 

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Input

The input consists of several test cases. The first line of each case contains two integers n<tex2html_verbatim_mark>(1n1000)<tex2html_verbatim_mark> and d<tex2html_verbatim_mark> , where n<tex2html_verbatim_mark> is the number of islands in the sea and d<tex2html_verbatim_mark> is the distance of coverage of the radar installation. This is followed by n<tex2html_verbatim_mark> 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

解题思路：

这个题目的大意为有一个坐标轴 在X轴上方是海 下方是陆地 X轴是海岸线，海上有N个小岛，现在要在海岸线上安装雷达，雷达的覆盖范围是一个以R为半径的圆，请用最少的雷达覆盖所有的小岛；当无法覆盖时 输出-1。我们可以算出 每个小岛能被覆盖的雷达的圆心，即以小岛为圆心 R为半径 作圆，该圆与X轴的交点：左交点为x-sqrt(R*R-y*y); 右交点为x+sqrt(R*R-y*y);按照 左交点 排序，如果i点的左交点在 当前雷达的右边 则需安装一个新雷达，更新雷达否则 如果 i点的右交点也在当前雷达的左边 则把当前雷达的圆心更新为该点的右交点

程序代码：

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const int N = 1001;
struct node
{
	double l,r;
}a[N];
bool fun(node a, node b)
{
	return a.l<b.l;
}
int main()
{
	int n,R,i,Case=0;
	double x,y,t;
	while(scanf("%d %d", &n, &R)&&n&&R)
	{
		int v=1;
		for(i=0; i<n; i++)
		{
			cin>>x>>y;
			a[i].l = x - sqrt(R*R-y*y);
			a[i].r = x + sqrt(R*R-y*y);
			if(y>R || R<=0 || y<0) 
				v=-1;
		}
		sort(a, a+n,fun);
		t = a[0].r;
		for(i=1; i<n && v!=-1; i++)
		if(a[i].l>t)
		{
			v++;
			t = a[i].r;
		}
		else if(a[i].r<t) t=a[i].r;
		printf("Case %d: %d\n",++Case, v);
	}
	return 0;
}