hdoj Radar Installation

Problem 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
#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;struct node{ float l,r,x,y;}d[1100];int cmp(node x,node y){return x.l<y.l;/*按照区域的最左端排序*/}int main(){int sum,t,m,n,flog=1,Case=1,i,num,j;while(scanf("%d%d",&m,&n),m||n){for(i=0;i<m;i++){scanf("%f%f",&d[i].x,&d[i].y);if(d[i].y>n) {flog=0;break;}else {d[i].l=d[i].x-sqrt(n*n-d[i].y*d[i].y);/*求出雷达的扫描区域*/d[i].r=d[i].x+sqrt(n*n-d[i].y*d[i].y);}}printf("Case %d: ",Case);if(flog==0) printf("-1\n");else {int sign=0;num=0;sort(d,d+m,cmp);num++;sign=d[0].r;/*这个很重要，sign始终标记雷达所能扫描的最右端*/for(i=1;i<m;i++)/*至少设置一个雷达，如果当前判断的区域达不到雷达最右端，加设一个雷达*/{if(d[i].r<sign)sign=d[i].r;else if(d[i].l>sign){num++;sign=d[i].r;}}printf("%d\n",num);}}return 0;}