Watering Grass UUV 1038 贪心

问题：

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

题意：有一块草坪，长为l，宽为w，在它的水平中心线上有n个位置可以安装喷水装置，各个位置上的喷水装置的覆盖范围为以它们自己的半径ri为圆。求出最少需要的喷水装置个数。

思路：贪心算法，如下图所示，将其转化成线区间来做。

代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{double x,y;double n,m;
} q[12220];
double zmh(node w,node t)
{return w.n<t.n;
}
int main()
{int a,b,c;while(~scanf("%d%d%d",&a,&b,&c)){double k=c/2.0;int j=0;for(int i=0; i<a; i++){scanf("%lf%lf",&q[i].x,&q[i].y);if(q[i].y*2>c)//直径小于长方形的宽度舍去{double z=sqrt(q[i].y*q[i].y-k*k);q[j].n=q[i].x-z;q[j++].m=q[i].x+z;}}sort(q,q+j,zmh);int ans=0;if(q[0].n>0){printf("-1\n");continue;}int flag;double l=0,r=0;int tag[11000]= {0},u;while(1){flag=0;for(int k=0; k<j; k++){if(tag[k]==0&&q[k].n<=l&&q[k].m>=r){flag=1;r=q[k].m;//更新右边的值if(r>=b)flag=2;u=k;}}if(flag==1||flag==2){ans=ans+1;l=r;tag[u]=1;}elsebreak;if(flag==2)break;}if(r<b)flag=0;if(flag==0)printf("-1\n");else printf("%d\n",ans);}return 0;
}