uva 10382

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

題意：

有一個草坪，在草坪中有一些噴水裝置，給出這個草坪的長和寬，和噴水裝置的位置和半徑，求最少放置多少個噴水裝置能使得草坪被全部覆蓋？

思路

這是一個區間覆蓋問題，可以參考算法競賽與入門經典P233，但是這個問題是圓形區域覆蓋，不好計算，因此我們把圓形區域轉換到草坪上來，就變成了矩形區域，這樣便於計算。

代碼：

#include<iostream>
#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;
}