【POJ1328】Radar Installation

Radar Installation

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu

Submit  Status  Practice  POJ 1328

Description

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d. 

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates. 
 
Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases. 

The input is terminated by a line containing pair of zeros 

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Sample Input

3 2
1 2
-3 1
2 1

1 2

0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

题解：找出每个点的区间，即圆与x轴的左右交点，按右交点从小到大排好，判断是否重叠，若重叠，说明能包含下一个点，雷达数不用增加，若不重叠，雷达数加一
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define N 1000000
struct node
{
	double xx,yy;//定义区间函数
}a[1000];
bool cmp(node a,node b)
{
	return a.yy<b.yy;
}
int main()
{
	int n,k=1,f;
	double r;
	while(scanf("%d%lf",&n,&r),n||r)
	{
		double x,y;
		f=1;
		for(int i=0;i<n;i++)
		{
		scanf("%lf%lf",&x,&y);
		if(f==0) continue;
		if(y>r)
		{
			f=0;
			continue;
			}
			a[i].xx=x-sqrt(r*r-y*y);
			a[i].yy=x+sqrt(r*r-y*y);
		}
		sort(a,a+n,cmp);
		if(f==0)
		{
			printf("Case %d: -1\n",k++);continue;
		}
		double temp=-N;
		int ans=0;
		for(int i=0;i<n;i++)
		{
			if(temp<a[i].xx)
			{
				ans++;
				temp=a[i].yy;
			}
		}
		printf("Case %d: %d\n",k++,ans);
	}
	return 0;
}
为啥下面这种解法不对！！我讨厌运行正确却不能AC的。。。

参考链接http://blog.youkuaiyun.com/lyy289065406/article/details/6642599

感觉和她方法一样 为啥不对

#include<cstdio>
#include<math.h>
#include<algorithm>
using namespace std;
double b[1000],c[1000];
struct node
{
	double x,y;
 } a[1000];
 bool cmp(node a,node b)
 {
 	return a.x<b.x;
 }
int main()
{
	int n,r,k=1;
	while(~scanf("%d%d",&n,&r))
	{
		if(n==0&&r==0)
		break;
		for(int i=0;i<n;i++)
		scanf("%d%d",&a[i].x,&a[i].y);
		sort(a,a+n,cmp);
		int f=0;
		for(int i=0;i<n;i++)
	{
		if(a[i].y>r)
		f=1;
	}
	if(f==1)
	{
		printf("Case %d: -1\n",k++);
		continue;
	}
	
		for(int i=0;i<n;i++)
		{
			b[i]=a[i].x-sqrt((double)(r*r-a[i].y*a[i].y));
			c[i]=a[i].x+sqrt((double)(r*r-a[i].y*a[i].y));
		}
		int t=1;
		double temp;
		for(int i=0,temp=c[0];i<n-1;i++)
		{
			if(b[i+1]>temp)
			{
				temp=c[i+1];
				t++;
			}
			else if(c[i+1]<temp)
			temp=c[i+1];	
		}
		printf("Case %d: %d\n",k++,t);
	}
	return 0;
}