[贪心]UVA10382 Watering Grass

Problem E
Watering Grass
Input: standard input
Output: standard output
Time Limit: 3 seconds

n sprinklers are installed in a horizontal strip of grass l meters long and w meters wide. Each sprinkler is installed at the horizontal center line of the strip. For each sprinkler we are given its position as the distance from the left end of the center line and its radius of operation.

What is the minimum number of sprinklers to turn on in order to water the entire strip of grass?

Input

Input consists of a number of cases. The first line for each case contains integer numbers n, l and w with n <= 10000. The next n lines contain two integers giving the position of a sprinkler and its radius of operation. (The picture above illustrates the first case from the sample input.)

 

Output

For each test case output the minimum number of sprinklers needed to water the entire strip of grass. If it is impossible to water the entire strip output -1.

Sample input

8 20 2

5 3

4 1

1 2

7 2

10 2

13 3

16 2

19 4

3 10 1

3 5

9 3

6 1

3 10 1

5 3

1 1

9 1

 

Sample Output

6

2

-1

(Regionals 2002 Warm-up Contest, Problem setter: Piotr Rudnicku)

 

题意：有一个草坪，在草坪中有一些喷水装置，给出这个草坪的长和宽，和喷水装置的位置和半径，求最少放置多少个喷水装置能使得草坪被全部覆盖？

思路：题目一看，就知道是区间覆盖问题，用贪心解决，因为是圆形区域覆盖不好计算，我们把圆形区域转换到草坪上来，就变成了矩形区域，这样就比较好计算了。

#include<iostream>
#include<algorithm>
#include<cmath>

using namespace std;

class Circle
{
public:
	double left,right;
}circle[10005];


int main()
{
	int num;
	double lenth,width;
	while(cin>>num>>lenth>>width)
		{
			int i,j,k=0;
			double pos,radius;
			for(i=0;i<num;i++)
				{
					cin>>pos>>radius;
					if(radius*2>=width)
						{
							double l,r;
							l=pos-sqrt(radius*radius-(width/2)*(width/2));
							r=pos+sqrt(radius*radius-(width/2)*(width/2));
							circle[k].left=l;
							circle[k++].right=r;
						}
				}
			double begin=0,end=lenth;
			double maxlen=0;
			int cnt=0;
			while(begin<end)
				{
					maxlen=0;
					for(i=0;i<k;i++)
						{
							if(circle[i].left<=begin&&circle[i].right>maxlen)
								{
									maxlen=circle[i].right;
								}
						}
					if(maxlen==begin)
						{
							cnt=-1;
							break;
						}
					cnt++;
					begin=maxlen;
				}
			cout<<cnt<<endl;
		}
	return 0;
}