题目链接：

啊哈哈，点我点我

题目：

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

Source

这道题目是贪心里面的区间选点问题。。贪心策略是：选区间的最右端的点.

思路：首先抽象出这个模型，以海岛为圆心，雷达距离为半径，求出在陆地上的区间，则雷达选在

这个区间之类那么必定能够扫描到这个海岛。。求出所有区间后，就转化成区间选点问题。。

还有就是代码中的那个标准end要用double，我wa了好久。。。

代码为：

#include<cstdio>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=1000+10;

struct Line
{
   double le,ri;
}line[maxn];

bool cmp(Line a,Line b)
{
   if(a.ri!=b.ri)
        return a.ri<b.ri;
   else
        return a.le>b.le;
}

int main()
{
    bool ok;
    int u,v,cas=1;
    int cnt;
	double End;
    int n,d;
    while(~scanf("%d%d",&n,&d))
    {
    	if(n==0&&d==0)   return 0;
        cnt=0;
        ok=false;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&u,&v);
            if(v>d)
                ok=true;
            else
            {
                line[i].le=(double)u-sqrt((double)(d*d-v*v));
                line[i].ri=(double)u+sqrt((double)(d*d-v*v));
            }
        }
        if(ok)
            printf("Case %d: -1\n",cas++);
        else
        {
           sort(line+1,line+1+n,cmp);
           cnt=0;
           End=-INF;
           for(int i=1;i<=n;i++)
           {
               if(line[i].le>End)
               {
                   cnt++;
                   End=line[i].ri;
               }
           }
           printf("Case %d: %d\n",cas++,cnt);
        }
    }
    return 0;
}