POJ-1328-Radar Installation-2013-12-07 01:49:28

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 45922 Accepted: 10252

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

Source

Beijing 2002

# include<stdio.h># include<string.h># include<math.h># include<iostream># include<algorithm>using namespace std;typedef struct Point{double left;double right;}p;int cmp(p x,p y){return x.left < y.left;}int main(){int i,n,k,r,x,y,flag,sum;double t;p point[1010];k = 0;while(~scanf("%d%d",&n,&r) && n!=0 || r!=0){flag = 1;for(i=0;i<n;i++){scanf("%d%d",&x,&y);if(abs(y)>r)flag = 0;point[i].left = x*1.0 - sqrt(r*r*1.0 - y*y*1.0);point[i].right = x*1.0 + sqrt(r*r*1.0 - y*y*1.0);}sort(point,point+n,cmp);sum = 1;t = point[0].right;for(i=1;i<n;i++){if(point[i].right<t){t = point[i].right;}else if(point[i].left>t){t = point[i].right;sum++;}}printf("Case %d: ",++k);if(flag==0)printf("-1\n");elseprintf("%d\n",sum);}return 0;}