解题思路：给出n个岛屿，n个岛屿的坐标分别为（a1,b1）,(a2,b2)-----(an,bn),雷达的覆盖半径为r

求所有的岛屿都被覆盖所需要的最少的雷达数目。

首先将岛屿坐标进行处理，因为雷达的位置在x轴上，所以我们设雷达的坐标为(x,0),对于任意一个岛屿p(a,b)，因为岛屿要满足在雷达的覆盖范围内，所以 (x-a)^2+b^2=r^2,解得

xmin=a-sqrt(r*r-b*b);//即为区间的左端点 xmax=a+sqrt(r*r-b*b);//即为区间的右端点

接下来区间选点即可  

                   ------------------------------tmp   

1                                                                       a[i]----------------------------b[i]     

2                             a[i]--------b[i]

3                                       a[i]-------------------- b[i]     

用tmp记录当前雷达坐标，将区间按左端点升序排序后，从左到右扫描，会出现 以上3种情况

1 当前tmp<a[i]，雷达无法覆盖到下一个区间，所以增加一个新的雷达，同时更新雷达的坐标为该区间的右端点(贪心，在越右边，越有可能覆盖到下一个区间),即为tmp=b[i]

2 当前b[i]<tmp，雷达无法覆盖该区间，但是该区间被包含在tmp所在区间内，所以不需要增加雷达，更新tmp的值即可 tmp=b[i]

3  该区间被雷达覆盖，不做处理。

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

#include<stdio.h>#include<string.h>#include<math.h>double a[1005],b[1005];void bubblesort(double a[],double b[],int n){	int i,j;	double t;	for(i=1;i<=n;i++)	{		for(j=i+1;j<=n;j++)		{			if(a[i]>a[j])			{				t=a[i];				a[i]=a[j];				a[j]=t;								t=b[i];				b[i]=b[j];				b[j]=t;							}		}	}}int main(){	int n,i,sum,flag,tag=0;	double x,y,r,tmp;	flag=1;	while(scanf("%d %lf",&n,&r)!=EOF&&(n||r))	{		tag=0;		for(i=1;i<=n;i++)		{			scanf("%lf %lf",&x,&y);						if(r>=fabs(y))			{			a[i]=x-sqrt(r*r-y*y);			b[i]=x+sqrt(r*r-y*y);		    }		    else		    tag=1;		    		}		if(tag)//不考虑r<0的情况也能通过 		printf("Case %d: -1\n",flag++);		else		{		  bubblesort(a,b,n);		  sum=1;		  tmp=b[1];		   for(i=2;i<=n;i++)		    {				if(a[i]>tmp)			{				tmp=b[i];				sum++;			}			else if(b[i]<tmp)			tmp=b[i]; 		    }		printf("Case %d: %d\n",flag++,sum);     	}		}}

POJ 1328 Radar Installation 【贪心 区间选点】