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.
                                                
                                                                       

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

题意为，在一个坐标系中，x轴代表海岸，x轴以上有n个点，代表着n个岛屿，在x轴上建雷达，一直雷达的覆盖范围为半径为d的圆，求在x轴上最少建多少个雷达，才能把全部的岛屿覆盖起来，如果不能覆盖，输出-1.

这题变相考区间不相交问题。以岛屿为圆心，以雷达的覆盖范围为半径画圆，与x轴交于两点，在这两点之间建雷达一定可以覆盖这个岛屿。这两点就是这个岛屿的可行区间，

岛屿之间的可行区间不相交就说明需要再建雷达，不能共用一个雷达。

注意精度问题，因为这错了好几次。

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
struct nde
{
    double x,y;
} a[1001];
bool cmp(nde x1,nde y1)
{
    return x1.y<y1.y;
}
int main()
{
    int l=0,flag,n,i,k;
    double r;
    while(~scanf("%d%lf",&n,&r)&&n&&r)
    {
        l++;
        flag=0;
        for(i=0; i<n; i++)
        scanf("%lf%lf",&a[i].x,&a[i].y);
        cout<<"Case "<<l<<": ";
        for(i=0; i<n; i++)
        {
            if(a[i].y>r)
            {
                flag=1;
                cout<<-1<<endl;
                break;
            }
            if(a[i].y==r)
                a[i].y=a[i].x;
            else
            {///将坐标转化为区间
                double t=sqrt(r*r-a[i].y*a[i].y);
                a[i].y=a[i].x+t;
                a[i].x=a[i].x-t;
            }
        }
        if(!flag)
        {///计算不相交区间个数
            k=1;
            sort(a,a+n,cmp);
            for(i=1; i<n; i++)
            {
                if(a[i].x>a[0].y)
                {
                    k++;
                    a[0].y=a[i].y;
                }
                else
                    continue;
            }
            cout<<k<<endl;
        }
    }
}