poj 1328 Radar Installation

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

Output

Sample Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

思路：把岛屿可能所在位置转化为区间

代码1：

#include<stdio.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;

struct node
{
	double x1,x2;
}p[1003];

bool cmp (node a,node b)
{
	return (a.x2<b.x2); 
}

int main ()
{
	int n,r;
	double x,y;
	int i,j,t=0,sum;
	
	while(cin>>n>>r)
	{
		if(n==0&&r==0) break;
		int flag=1;
		t++;
		
		for(i=0;i<n;i++)
		{
			cin>>x>>y;
			if(y>r || y<0)  flag=0; 
			p[i].x1=x-sqrt(r*r-y*y);
			p[i].x2=x+sqrt(r*r-y*y);
		}
		
		if(flag==0) 
		{
			cout<<"Case "<<t<<":"<<" "<<"-1"<<endl;
			continue;
		}
		
		sort(p,p+n,cmp);
		sum=1;
		double temp=p[0].x2;
		for(i=1;i<n;i++)
		{
			if(p[i].x1>temp)
			{
				temp=p[i].x2;
				sum++;
			}
		}
		cout<<"Case "<<t<<":"<<" "<<sum<<endl;
	}
	return 0;
}

#include<stdio.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;

struct node
{
	double x1,x2;
}p[1003];

bool cmp (node a,node b)
{
	return (a.x1<b.x1); 
}

int main ()
{
	int n,r;
	double x,y;
	int i,j,t=0,sum;
	
	while(cin>>n>>r)
	{
		if(n==0&&r==0) break;
		int flag=1;
		t++;
		
		for(i=0;i<n;i++)
		{
			cin>>x>>y;
			if(y>r || y<0)  flag=0; 
			p[i].x1=x-sqrt(r*r-y*y);
			p[i].x2=x+sqrt(r*r-y*y);
		}
		
		if(flag==0) 
		{
			cout<<"Case "<<t<<":"<<" "<<"-1"<<endl;
			continue;
		}
		
		sort(p,p+n,cmp);
		sum=1;
		double temp=p[0].x2;
		
		for(i=1;i<n;i++)
		{
			if(p[i].x2<temp) temp=p[i].x2;
			else
				if(p[i].x1>temp)
				{
					temp=p[i].x2;
					sum++;
				}
		}
		
		cout<<"Case "<<t<<":"<<" "<<sum<<endl;
	}
	return 0;
}

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poj 1328 Radar Installation

原文：http://blog.csdn.net/fyxz1314/article/details/37882523