source

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

區(qū)間選點(diǎn)：數(shù)軸上有n個(gè)閉區(qū)間[ai,bi]积仗。取盡量少的點(diǎn)，使得每個(gè)區(qū)間內(nèi)都至少有一個(gè)點(diǎn)（不同區(qū)間內(nèi)含的點(diǎn)可以是同一個(gè)）湃番。
題意：給出n個(gè)島，和雷達(dá)能覆蓋的最大半徑胚想，島嶼分布在水平線以上潘拨，雷達(dá)安裝在水平線上，求所需的最少的雷達(dá)數(shù)环鲤。若存在雷達(dá)不能覆蓋所有島嶼的情況，輸出-1嗅辣。
題解 :求出每個(gè)島嶼能夠被雷達(dá)覆蓋的雷達(dá)安裝區(qū)間撼泛，然后就等價(jià)于區(qū)間選點(diǎn)問題。

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
struct section{

   double x,y;

} arr[1010];
bool cmp(const section & a,const section & b)
{
        return (a.y<b.y)||((a.y==b.y)&&(a.x>b.x));
}
int main()
{
    int n,cnt=1,flag;
    double d,x0,y0;
    while(scanf("%d%lf",&n,&d)&&(n||d)){
            int sum=1;
            flag=0;
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&x0,&y0);
            if(y0>d) {
                flag=1;
                continue;
            }
            double sqr=sqrt(d*d-y0*y0);
            arr[i].x=x0-sqr;
            arr[i].y=x0+sqr;
        }
        if(flag){
            printf("Case %d: -1\n",cnt);
        }
        else
        {
            sort(arr,arr+n,cmp);
            double dy=arr[0].y;
            for(int i=1;i<n;i++)
            {
                if(arr[i].x>dy)
                {
                    sum++;
                    dy=arr[i].y;
                }
            }
            printf("Case %d: %d\n",cnt,sum);
        }
        cnt++;
    }
}