Radar Installation

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 86867 Accepted: 19450

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：

很容易就看出这是一道贪心题，不过需要思考的是怎样去转化一下可以轻松实现出来，以岛屿为圆心，以雷达的半径为半径画圆，截得的数轴的线段就是雷达要覆盖该岛可以放的位置，这样就成了线段求贪心了，求有几段不重叠的线段就可以了。注意：精度！！

#include<iostream>#include<math.h>#include<algorithm>using namespace std;struct point{    double r,l;};bool cmp(const struct point a,const struct point b){    return a.l<b.l;}int main(){    int n,d;    int j=0;    while(cin>>n>>d&&(n||d)){        point p[1005];        int f=1;        for(int i=0;i<n;i++){            int a,b;            cin>>a>>b;            if(b>d)f=0;            double an=sqrt(d*d*1.0-b*b*1.0);            p[i].r=a*1.0-an;//左端点，            p[i].l=a*1.0+an;//右端点，单词弄错了，只好将错就错了。        }        sort(p,p+n,cmp);        int ans=0;        double ls;        for(int i=0;i<n;i++){            if(i==0){                ls=p[i].l;                ans++;            }            else{                if(p[i].r>ls){                    ls=p[i].l;                    ans++;                }            }        }        j++;        if(f==1)cout<<"Case "<<j<<": "<<ans<<endl;        else cout<<"Case "<<j<<": "<<"-1"<<endl;    }}