Radar Installation(贪心)

Radar Installation

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总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

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

输入

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

输出

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.

样例输入

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

样例输出

Case 1: 2
Case 2: 1

1.思路：

    给出一个点，然后根据半径找到这个点左右区间。计算出所有的区间，然后将所有的区间按照起始点递增的顺序排序，最后将区间合并处理。

    1.1如果后一个的左区间大于前一个的右区间，雷达数++，更新区间为当前区间

    1.2如果后一个的右区间小于前一个的右区间，更新右区间即可，因为可以覆盖。

2.为什么以左区间为临界点？？？(正如下面图片中所显示，第四条线段的起点并不能覆盖前三条，所以应该在第三条的起点处放置雷达。这也就是，为什么用左端点排序的原因)

   

 

#include<iostream>
#include<algorithm>
#include<map>
#include<string>
#include<vector>
#include<cmath>
using namespace std;
struct node
{
	double zuo,you;
	bool operator<(const node &no)const
	{
		return zuo<no.zuo;
	}
}no[1006];

int main()
{
	int n;
	double r;
	int cnt=0;
	while(scanf("%d %lf",&n,&r)!=EOF)
	{
		cnt++;
		int fz=0;
		if(n==0||r==0) break;
		for(int i=0;i<n;i++)
		{
			double a,b;
			cin>>a>>b;
			if(b>r)
			{
				fz=1;
				continue;
			}
			no[i].zuo=a-sqrt(r*r-b*b);
			no[i].you=a+sqrt(r*r-b*b);
		}
		if(fz)
		{
			printf("Case %d: -1\n",cnt);
		}
		else
		{
			sort(no,no+n);
			int sum=1;
			double left=no[0].zuo,right=no[0].you;
			for(int i=1;i<n;i++)
			{
			
				if(no[i].zuo>right)//并不能将前一个区间覆盖，此时雷达数加一，区间变为当前区间 
				{
					
					sum++;
					left=no[i].zuo;
					right=no[i].you;
				}
				else if(no[i].you<right) //可以将前一个区间覆盖，右区间更新即可 
				       right=no[i].you;
				
			}
			printf("Case %d: %d\n",cnt,sum);
		}
	}
	return 0;
}