贪心（区间覆盖）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 101412		Accepted: 22561

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

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

题意：地图的x轴的上方为海，下方为陆地，海中有n个小岛，坐标为(isl[i].x,isl[i].y)。有一种雷达，能探测到的范围为以d为半径的圆。问海岸线上至少造多少雷达可以把所有的小岛都包含在内。注意雷达是建在海岸线上的，也就是x轴上的。

思路：贪心，从左到右建立雷达，要尽量多地覆盖岛屿。以岛屿为圆心，以d为半径画圆，如果画出的圆与x轴没有交点，则不能实现。存在交点的话，计算出第i个岛屿与x轴的交点坐标保存在结构体数组rad[i].sta与rad[i].end中。对结构体数组排序，按照rad[i].end进行升序排列，然后一次从左到右找雷达。对于rad[i].end为当前最右边的左坐标，对于下一个岛屿，如果rad[j].sta<ran[i].end，则不需要建立新的雷达，需要更新下一个岛屿。否则需要建立新的雷达。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
struct island
{
	int x;
	int y;
}isl[1005];//小岛坐标 
struct range
{
	float sta;
	float end;
}ran[1005];//以小岛为圆心d为半径做圆与x轴相交区域 
bool cmp(range a,range b)
{
	return a.end<b.end;
}
int main()
{
	int n,d,t=1;
	while(cin>>n>>d&&n!=0)
	{
		int i,j,maxy=0;
		for(i=0;i<n;i++)
		{
			cin>>isl[i].x>>isl[i].y;
			if(isl[i].y>maxy) maxy=isl[i].y;
		}
		cout<<"Case "<<t++<<": ";
		if(maxy>d||d<0)
		{
			cout<<"-1"<<endl;
			continue;
		}
		float len;
		for(int i=0;i<n;i++)//求小岛对应雷达的可能覆盖范围 
		{
			len=sqrt(1.0*d*d-isl[i].y*isl[i].y);
			ran[i].sta=isl[i].x-len;
			ran[i].end=isl[i].x+len;
		}
		sort(ran,ran+n,cmp);//根据ran的end值进行排序 
		int ans=0;
		bool vis[1005];memset(vis,false,sizeof(vis));
		for(int i=0;i<n;i++)
		{//类似的活动选择 
			if(!vis[i])
			{
				vis[i]=true;
				for(j=0;j<n;j++)
				{
					if(!vis[j]&&ran[j].sta<=ran[i].end)
					vis[j]=true;
				}
				ans++;
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}