poj 1328 Radar Installation

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 87187 Accepted: 19540

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

思路详见代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;int n;double d;double data[1005][2];struct Node{double l,r;}dir[1005];int cmp(Node a,Node b){return a.l<b.l;}int cal()//计算能扫到每个岛的区间{int index=0;for(int i=0;i<n;i++){if(data[i][1]>d)//有一个岛屿扫不到，结果直接 -1return 0;double dis=sqrt(d*d-data[i][1]*data[i][1]);dir[index].l=data[i][0]-dis;dir[index].r=data[i][0]+dis;index++;}}int main(){int cc=0;int ans;while(scanf("%d%lf",&n,&d),n+d){ans=0;for(int i=0;i<n;i++)scanf("%lf%lf",&data[i][0],&data[i][1]);if(cal()){sort(dir,dir+n,cmp);double ll,rr;ll=dir[0].l;rr=dir[0].r;for(int i=1;i<n;i++)//对所求的区间做交集，最后看看有几个不交叉的集合，就是几个雷达。 {if(dir[i].l>rr){ll=dir[i].l;//注意比较对象的改变 rr=dir[i].r;ans++;}else if(dir[i].l<=rr){ll=dir[i].l;if(dir[i].r<=rr) rr=dir[i].r;}}printf("Case %d: %d\n",++cc,++ans);}else{printf("Case %d: -1\n",++cc);} }return 0;}