UVA 10382 Watering Grass（区间嵌套）

最新推荐文章于 2019-07-23 18:57:57 发布

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最新推荐文章于 2019-07-23 18:57:57 发布

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分类专栏：贪心文章标签： uva UVA 区间嵌套贪心

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本文探讨了如何使用贪心算法解决草坪灌溉问题，通过精确计算每个喷头的有效覆盖范围和位置，确保整个草坪得到充分灌溉。算法考虑了喷头的位置、覆盖范围以及草坪的宽度，从而确定最少数量的喷头配置，实现高效灌溉。

Problem E
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 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

-1

转化为区间嵌套，用圆和矩形的边形成的弦区间，贪心求之，注意精度问题

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstdlib>
using namespace std;
struct data{
	double x,y;
}f[10010];
#define eps 1e-7
inline int ifcmp(double x){//精度
    if (x > -eps) return 1;
    else return -1;
}
bool cmp(data a,data b)
{
	return a.x<b.x||(a.x==b.x&&a.y>b.y);
}
int main()
{
	int n,l,w,i,j,k,a,b;
	while(scanf("%d%d%d",&n,&l,&w)!=EOF)
	{
		for(i=0,j=0;i<n;i++)
		{
			scanf("%d%d",&a,&b);
			if(2*b<w)	
				continue;
			double tem;
			tem=sqrt((double)b*b-(double)w*w/4.0);//半弦
			f[j].x=a-tem;
			f[j].y=a+tem;
			j++;
		}
		sort(f,f+j,cmp);
		int pre=0,pos,sum=0,max_j;
		sum=1;
		i=1;
		max_j=0;
		while(true)
		{
			bool flag=false;
			while(ifcmp(f[pre].y-f[i].x)>0&&i<j)
			{
				flag=true;
				if(ifcmp(f[i].y-f[max_j].y)>0)
					max_j=i;
				i++;
			}
			if(!flag||(i>j&&ifcmp(f[max_j].y-(double)l)<0))
			{
				sum=-1;
				break;
			}
			pre=max_j;
			sum++;
			if(ifcmp(f[max_j].y-(double)l)>0)
				break;
		}
		printf("%d\n",sum);
	}
}