Radar installation (greedy, can be converted to this summer holiday not AC type)

Radar installationtime limit:2000/1000ms (java/other) Memory limit:20000/10000k (Java/other) total submission (s): 54 Accepted Submission (s): 28Problem Descriptionassume The coasting is a infinite straight line. Side of coasting, sea in the other. Each of small island was a point locating in the sea side. and any radar installation, locating on the coasting, can only cover D-distance, so a island in the sea can is 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 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 a is represented by its X-y coordinates.

Figure A Sample Input of Radar installations

Inputthe input consists of several test cases. The first line of all case contains-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 was followed by n lines each containing and 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

Outputfor each test case output one line consisting of the "test case" number followed by the minimal number of radar instal Lations needed. "-1" installation means no solution for this case.

Sample INPUT3 2 1 2-3 1 2 1 1 2 0 2 0 0

Sample outputcase 1:2 Case 2:1: First conversion to interval points, then sorting, the interval to find points; code:

1#include <stdio.h>2#include <math.h>3#include <algorithm>4 using namespacestd;5 structnode{6     Doubles,e;7 };8 intn,d,k;9Node area[1010];Ten intCMP (Node A,node b) { One     returna.e<B.E; A } - intChangeintXinty) { -     if(y>d)return 0; the     DoubleA,B,M=SQRT (d*d-y*y); -a=x-m;b=x+m; -area[k].s=a;area[k].e=b;k++; -     return 1; + } - intMain () {intT,x,y,flot,temp,num,l=0; +      while(SCANF ("%d%d", &n,&d), n| | d) {k=0; flot=1; temp=0; num=1; l++; A              for(intI=0; i<n;i++){ atscanf"%d%d",&x,&y); -t=Change (x, y); -                 if(!t) flot=0; -             } -Sort (area,area+k,cmp); -              for(intI=0; i<k;i++){ in                 if(AREA[I].S&GT;AREA[TEMP].E) temp=i,num++; -             } toprintf"Case %d:%d\n", l,num); +     } -     return 0; the}

Radar installation (greedy, can be converted to this summer holiday not AC type)