POJ 1328Radar Installation

Radar Installation Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu Submit  Status  Practice  POJ 1328

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

#include<stdio.h> #include<cstring> #include<cmath> #include<algorithm> using namespace std; struct node { double x,y; }a[1010]; struct range { double l,r; }s[1010]; bool cmp(range A,range B) { return A.l<B.l; } int main() { int n,d,i; int k=1; while(scanf("%d%d",&n,&d)!=EOF&&n!=0&&d!=0) { int sum=1; for(i=0;i<n;i++) { scanf("%lf%lf",&a[i].x,&a[i].y); s[i].l=a[i].x-sqrt(d*d-a[i].y*a[i].y); s[i].r=a[i].x+sqrt(d*d-a[i].y*a[i].y); if(a[i].y>d||d<0||a[i].y<0) sum=-1; } sort(s,s+n,cmp); double m=s[0].r; for(i=1;i<n&&sum!=-1;i++) { if(m<s[i].l) { sum++; m=s[i].r; } else if(m>s[i].r) m=s[i].r; } printf("Case %d: %d\n",k++,sum); } }