区间选点_Radar Installation

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

题意：给出n个岛，和雷达能覆盖的最大半径，岛屿分布在水平线以上，雷达安装在水平线上，求所需的最少的雷达数。若存在雷达不能覆盖所有岛屿的情况，输出-1。
题解 :求出每个岛屿能够被雷达覆盖的雷达安装区间，然后就等价于区间选点问题。

#include
#include
#include
using namespace std;
struct section{

   double x,y;

} arr[1010];
bool cmp(const section & a,const section & b)
{
        return (a.yb.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;id) {
                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;idy)
                {
                    sum++;
                    dy=arr[i].y;
                }
            }
            printf("Case %d: %d\n",cnt,sum);
        }
        cnt++;
    }
}