Radar Installation

                                    Radar Installation

                                                        Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Problem 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

 

 题意：p 点表示 岛屿，x轴是海岸线，在轴上建立雷达装置，d 表示每个装置的覆盖半径， 问最少需要多少雷达装置可以把所有的岛屿覆盖，如果有无法覆盖的就输出“-1”；

思路：先计算出在x轴上能覆盖每个岛屿的范围（L，R），然后以 L 排序，最后根据 条件排序即可；详见代码;

源代码：

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=1005;
struct Line
{
    double l,r;
}line[maxn];
bool cmp(Line a,Line b)
{
    return a.l<b.l;
}
int main()
{
    int n,d;
    int x,y,i;
    bool yn;
    int icase=1;
    while(cin>>n>>d)
    {
        yn=true;
        int c=0;
        if(n==0&&d==0)
            break;
        for(i=0;i<n;i++)
        {
            cin>>x>>y;
            if(yn==false) continue;
            if(y>d) yn=false;
            else
            {
                line[i].l=(double)x-sqrt((double)d*d-y*y);
                line[i].r=(double)x+sqrt((double)d*d-y*y);
            }
        }
        if(yn==false)
        {
            cout<<"Case "<<icase++<<": -1"<<endl;
            continue;
        }
        sort(line,line+n,cmp);
        c++;
        double now=line[0].r;
        for(i=1;i<n;i++)                            ///计算条件
        {
            if(line[i].r<now)
                now=line[i].r;                      ///这点很重要！容易忘
            else if(now<line[i].l)                  ///关键条件
            {
                now=line[i].r;                      
                c++;
            }
        }
        cout<<"Case "<<icase++<<": "<<c<<endl;
    }
    return 0;
}