POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 50740 Accepted: 11394

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

将点转化为区间，以区间右端点进行排序。

AC代码如下：

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;struct H{    double l,r;}a[1005];bool cmp(H a,H b){    return a.r<b.r;}int main(){    int n,r,bj;    double h,z;    int i,j,cas=0;    while(cin>>n>>r,n!=0&&r!=0)    {        bj=0;        cas++;    for(i=0;i<n;i++)    {       cin>>h>>z;       if(z>r)       {           bj=1;       }       a[i].l = h - sqrt(r*r-z*z);       a[i].r = h + sqrt(r*r-z*z);    }    if(bj)    {cout<<"Case "<<cas<<": "<<"-1"<<endl;continue;}    sort(a,a+n,cmp);    int ans=1;    double L,R;    L=a[0].l;    R=a[0].r;    for(i=1;i<n;i++)    {        if(a[i].l>R)//想清楚这一步，就不难了        {            ans++;            R=a[i].r;        }    }    cout<<"Case "<<cas<<": "<<ans<<endl;    }    return 0;}