POJ-1328

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 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 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：

#include<iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <algorithm>
//#include<bits/stdc++.h>
using namespace std;

struct node{
    double le,ri;
}a[2010];

int cmp(node a,node b){
    return a.le<b.le;
}

int x[2010],y[2010];

int main(){
    ios::sync_with_stdio(false);
    int ans=0;
    int n,d,num,flag;
    double z,temp;
    while(cin>>n>>d){
        if(n==0&&d==0) break;
        ans++;
        flag=0;
        for(int i=0;i<n;i++){
            cin>>x[i]>>y[i];
            if(y[i]>d)
            flag=1;
        }
        if(flag){
            cout<<"Case "<<ans<<": -1"<<endl;
        }
        else{
            for(int i=0;i<n;i++){
                z=sqrt(d*d*1.0-y[i]*y[i]*1.0);
                a[i].le=(double)x[i]-z;
                a[i].ri=(double)x[i]+z;
            }
            sort(a,a+n,cmp);
            temp=-100000000;
            num=0;
            for(int i=0;i<n;i++){
                if(a[i].le>temp){
                    num++;
                    temp=a[i].ri;
                }else if(a[i].ri<temp){
                    temp=a[i].ri;
                }
            }
            cout<<"Case "<<ans<<": "<<num<<endl;
        }
    }
    return 0;
}