Radar Installations

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

代码：

#include<iostream>#include<cmath>#include<algorithm>using namespace std;struct island{double x;double y;bool operator <(island& a){return (y<a.y);}}pos[1001],sortpos[1001];int main(){int n,d,i,x,j=0;bool work;double current;cin>>n>>d;while(!(n==0&&d==0)){j++;work=false;for(i=0;i<n;i++){cin>>pos[i].x>>pos[i].y;if(pos[i].y>d)work=true;if(work==false){sortpos[i].x=double(pos[i].x-sqrt(d*d-pos[i].y*pos[i].y));sortpos[i].y=double(pos[i].x+sqrt(d*d-pos[i].y*pos[i].y));}}if(d<=0||work==true)          // d<=0,y<=0cout<<"Case"<<" "<<j<<": "<<-1<<endl;else{sort(sortpos,sortpos+n);if(n!=1){current=sortpos[0].y;x=1;for(i=1;i<n;i++){    if(sortpos[i].x>current){x++;current=sortpos[i].y;}else{if(sortpos[i].y<current)current=sortpos[i].y;}}}elsex=1;cout<<"Case"<<" "<<j<<": "<<x<<endl;//Case的首字母一定要大写}cin>>n>>d;}return 0;}