UVA 10382 Watering Grass (贪心）

原题：

      nsprinklers are installed in a horizontal strip of grassl
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
andw
with
n ≤ 10000. The nextn
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
62
-1
4000

思路：

 
    将圆的覆盖范围转化成长度端点，遍历一边即可，判断是否有间隙或无法达到最右端。注意sqrt时的精度处理和圆直径小于等于宽时无法覆盖。

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <deque>
#include <string>
#include <cmath>
#include <vector>
#include <utility>
#include <set>
#include <map>
#include <sstream>
#include <climits>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define pi acos(-1.0)
#define INF 2147483647
using namespace std;
typedef long long ll;
typedef pair <int,int > PP;
int N;
struct length
{
double le,ri;
}len[10050];
bool cmp(const length&a,const length &b)
{
return a.le<b.le;
}
int main()
{
int n;
double l,w;
while(scanf("%d%lf%lf",&n,&l,&w)!=EOF)
{
for(int i=0;i<=10000;i++)
{
len[i].le=0;
len[i].ri=0;
}
for(int i=0;i<n;i++)
{
double o,r,dis;
scanf("%lf%lf",&o,&r);
r=r*2;
if(r<=w)
dis=0;
else
dis=sqrt(r*r-w*w+0.5)/2;
//printf("%lf\n",dis);
len[i].le=o-dis;
len[i].ri=o+dis;
}
sort(len,len+n,cmp);
// for(int i=0;i<n;i++)
// cout<<len[i].le<<" "<<len[i].ri<<endl;
int res=0,j=0;
bool no=false;
for(double i=0.0;i<=l;)
{
double max_=-1;
for(;len[j].le<=i&&j<n;j++)
{
if(max_<len[j].ri)
max_=len[j].ri;
}
// cout<<j<<endl;
if(max_==-1)
{
no=true;
break;
}
res++;
i=max_;
//cout<<i<<" "<<max_<<endl;
}
if(no)
printf("-1\n");
else
printf("%d\n",res);
}
return 0;
}