Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 89640		Accepted: 20135

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

题意：

有n座岛，在x轴上最少建立多少半径为d的雷达可以覆盖所有的岛。

贪心。

首先以岛为圆心，d为半径求出x轴上符合的区间，再将区间按左到右的顺序排序，若当前点的左端点大于之前最大的右端点，则++。更新最右端点。

AC代码：

 1 #include<iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <string.h> 5 #include <math.h> 6 #include <algorithm> 7 //#include<bits/stdc++.h> 8 using namespace std; 9 10 struct node{11     double le,ri;12 }a[2010];13 14 int cmp(node a,node b){15     return a.le<b.le;16 }17 18 int x[2010],y[2010];19 20 int main(){21     ios::sync_with_stdio(false);22     int ans=0;23     int n,d,num,flag;24     double z,temp;25     while(cin>>n>>d){26         if(n==0&&d==0) break;27         ans++;28         flag=0;29         for(int i=0;i<n;i++){30             cin>>x[i]>>y[i];31             if(y[i]>d)32             flag=1;33         }34         if(flag){35             cout<<"Case "<<ans<<": -1"<<endl;36         }37         else{38             for(int i=0;i<n;i++){39                 z=sqrt(d*d*1.0-y[i]*y[i]*1.0);40                 a[i].le=(double)x[i]-z;41                 a[i].ri=(double)x[i]+z;42             }43             sort(a,a+n,cmp);44             temp=-100000000;45             num=0;46             for(int i=0;i<n;i++){47                 if(a[i].le>temp){48                     num++;49                     temp=a[i].ri;50                 }else if(a[i].ri<temp){51                     temp=a[i].ri;52                 }53             }54             cout<<"Case "<<ans<<": "<<num<<endl;55         }56     }57     return 0;58 } 

POJ-1328