Q

Q - Radar Installation

POJ - 1328 

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

思路： 转换看图的方式，。。。

区间选点问题，  多个区间，问至少要几个点才可以 在每个区间里都有点存在

代码

#include<stdio.h>#include<algorithm>#include<math.h>typedef struct{double left,right;}data;data a[1000+10];bool cmp(data a,data b){if(a.right!=b.right) return a.right<b.right;return a.left>b.left; }  using namespace std; int main() { int n,d;int p=1; while(scanf("%d%d",&n,&d)&&(n!=0||d!=0)) { int l=1; for(int i=0;i<n;i++) { int x,y; double k; scanf("%d%d",&x,&y); k=d*d*1.0-y*y*1.0;   //  转化为 以岛屿为源点 d为半径的圆 与海岸线的交线段 if(k<0||d<0) l=0; else if(l) a[i].left=x-sqrt(k); a[i].right=x+sqrt(k); } //putchar('\n'); if(!l) {  printf("Case %d: -1\n",p++); continue;}sort(a,a+n,cmp);double rightt=a[0].right;int sum=1;for(int i=1;i<n;i++)    //区间选点问题{if(rightt<a[i].left){sum++;rightt=a[i].right;}}printf("Case %d: %d\n",p++,sum); } }