POJ 1328 Radar Installation 贪心

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
<p>Case 2: 1</p>

题意 ：有n个小岛 并给出海岸线上雷达覆盖的半径d，雷达只能放置于海岸线上（x轴），给出每个小岛的坐标，问至少要多少个雷达才能把小岛全部覆盖。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
struct node{
	double l,r;
}q[1002];
bool cmp(node a,node b)
{
	return a.r<b.r;
}
int main()
{
	int m,n,a,b,i,t,k=0,s;
	while(cin>>m>>n)
	{
		k++;
		t=0;      
		if(m==0&&n==0)
			break;
		s=1;
		for(i=0;i<m;i++)
		{
			cin>>a>>b;
			if(b>n)                        //如果某岛y值大于半径 一定不能覆盖
				t=1;               
			q[i].l=a-sqrt((double)n*n-b*b);
			q[i].r=a+sqrt((double)n*n-b*b);        //求出每个岛要被覆盖 能放置的雷达区域
		}
		if(t==1)
		{
			cout<<"Case "<<k<<": -1"<<endl;
			continue;
		}
		sort(q,q+m,cmp);                //关于右端排序
		double ld=q[0].r;
		for(i=1;i<m;i++)
		{
			if(q[i].l>ld)                
			{
				s++;
				ld=q[i].r;
			}
			else if(q[i].r<ld)
				ld=q[i].r;
		}
		cout<<"Case "<<k<<": "<<s<<endl;
	}
	return 0;
}

将每个小岛的能安置的雷达区间求出来，然后对每个区间的右端点按从小到大排序，将第一个雷达放在第一个区间的右端点，然后对下一个区间进行判断：
1.如果下一个区间的左端点在雷达的右边，那么雷达一定不能覆盖到下一个岛屿，雷达数加1，将新雷达放置在下一个区间的右端点
2.如果下一个区间的右端点在雷达的左边，那么雷达能够覆盖，将雷达放置到下一个区间的右端点
3.如果下一个区间的左端点在雷达的左边，右端点在雷达的右边，即两个区间相交，那么雷达能覆盖下一个岛屿，雷达位置不变
注意用double处理