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。
Radar Installation
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 55518 | Accepted: 12502 |
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
#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);
}
}
}