## Water spray （ Two ）

The time limit ：3000 ms  |  Memory limit ：65535 KB
difficulty ：4

describe
There's a lawn , Transverse length w, The longitudinal length is h, It is installed at different positions on its transverse center line n(n<=10000) A dot sprinkler , Every sprinkler i The effect of spraying water is to make it center with a radius of Ri All the circles are wetted . Please choose as few sprinklers as possible from the ones given , Wet the whole lawn .

Input
Enter a positive integer in the first line N Expressing common ownership n Test data .
The first line of each set of test data has three integers n,w,h,n Expressing common ownership n A sprinkler ,w Indicates the transverse length of the lawn ,h Represents the longitudinal length of the lawn .
And then n That's ok , Both have two integers xi and ri,xi It means the first one i The abscissa of a sprinkler （ On the far left is 0）,ri Represents the radius of the circle that the sprinkler can cover .
Output
Each group of test data output a positive integer , Indicates how many sprinklers are needed , Each output occupies a separate line .
If there's no way to wet the whole lawn , Please export 0.
The sample input
```2
2 8 6
1 1
4 5
2 10 6
4 5
6 5 Explain ： It belongs to the greedy optimal solution problem  Code ```
```#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
struct T
{
int a,b;
}c[];
int cmp(T n,T m)
{
if(n.a<m.a)
return ;
if(n.a==m.a && n.b>m.b)
return ;
return ;
}
int main()
{
int i,m,n,k=;
int w,h,x,r;
scanf("%d",&m);
while(m--)
{int k=; int max=-;
scanf("%d %d %d",&n,&w,&h);
for(i=;i<n;i++)
{
scanf("%d %d",&x,&r);
if(*r>h)// Get rid of the ones that don't meet the conditions ,
{
c[k].a=x-sqrt(r*r-h*h/);
if(c[k].a<)
c[k].a=;
c[k].b=sqrt(r*r-h*h/)+x;
if(c[k].b>w)
c[k].b=w;
if(c[k].b>max)
max=c[k].b;
k++;// Recalculate the number of groups in the array
}
}
sort(c,c+k,cmp);// Quick line up , Prevent overtime
if(c[].a!= || max!=w)
printf("0\n");
else
{ int count=;
int f=;
int start=,last=;
while(start!=w)
{
for(i=f; i<k; i++)
if(c[i].a<=start && c[i].b>last)
{
last=c[i].b;
f=i+;
}
if(last==start)// There is no solution to this situation
{
count=;
break;
}
start=last;// Redefine the boundary
count++;
}
cout<<count<<endl;
}
}
return ;
}```

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