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 andw 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 numbersn, 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
思路分析：

读入数据，并计算其左右端点：

左：x-sqrt（r*r-（w/2）*（w/2））;

右：x-sqrt（r*r+（w/2）*（w/2））;

2.对其左端点进行从大到小排序

3.从左到右依次处理区间，注意要考虑的情况：

（1）若最小左端点不满足，直接得出不可以完全覆盖；

（2）若中间存在间断，得不可以完全覆盖；

（3）若到最后也不能满足右端点达到草坪的长度，得不可以完全覆盖。

代码如下：

1. #include<stdio.h>
   #include<math.h>
   #include<string.h>
   #include<algorithm>
   using namespace std;
   int n,l,w;
   struct node
   {
       double x,xx,yy;
   } a[120000];
   bool cmp(node u,node v)
   {
       return u.xx<v.xx;
   }
   int book[12000];
   int main()
   {
       while(~scanf("%d %d %d",&n,&l,&w))
       {
           double b,c;
           int j=0;
           for(int i=0; i<n; i++)
           {
               scanf("%lf %lf",&b,&c);
               if(c*2>w)        //舍去无效的喷水装置，并求有效的喷水装置的左右端点
               {
                   a[j].x=b;
                   double r=sqrt(c*c-w*w*1.0/4);
                   a[j].xx=b-r;    //左端点
                   a[j++].yy=b+r;  //右端点
               }
           }
           sort(a,a+j,cmp);        //对左端点进行排序
           if(a[0].xx>0)           //如果最小的左端点小于0则不可能完全覆盖
           {
               printf("-1\n");     
               continue;
           }
           double k=0,h=0;
           int sum=0,flag,q;
           memset(book,0,sizeof(book)); 
           while(1)
           {
               flag=0;
               for(int i=0; i<j; i++)        
               {
                   if(a[i].xx<=h&&a[i].yy>=k&&book[i]==0)//遍历，去更新满足左端点的最大右端点
                   {
                       flag=1;
                       k=a[i].yy;
                       if(k>=l)
                           flag=2;
                       q=i;
                   }
               }
               if(flag==1||flag==2)
               {
                   sum++;
                   h=k;
                   book[q]=1;    //对于用过的点进行标记
               }
               else
                   break;        //找不到满足的情况直接跳出循环，证明不能完全覆盖
               if(flag==2)       //已经满足完全覆盖，跳出循环
                   break;    
           }
           if(k<l)
               flag=0;
           if(flag==0)
               printf("-1\n");
           else
               printf("%d\n",sum);
       }
       return 0;
   }