POJ 1328 Radar Installation 贪心算法

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

Source

#include<cstdio>
#include<set>
#include<map>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
#include<string>
#include<cmath>
using namespace std;
typedef long long LL;
#define MAXN 1003

/*
对于每个点可以在坐标轴上 得出能覆盖到该点的圆心范围，从而转化为x轴上的很多线段
选取一定数目的点，让所有的线段都包含至少一个点
先按结尾端点排序，然后尽量将雷达分布在右端（这样能尽可能和多个线段重叠）
*/
struct node
{
    double beg,end;
}a[MAXN];
int n,d;
void cal(double x,double y,double &beg,double &end)//要求y<=d
{
    double r = (double)d;
    beg = x - sqrt(r*r-y*y);
    end = x + sqrt(r*r-y*y);
}
bool cmp(node a,node b)
{
    return a.end<b.end;
}
int main()
{
    int cas = 1;
    while(scanf("%d%d",&n,&d),n+d)
    {
        double x,y;
        bool f = false;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&x,&y);
            if(!f&&y<=d)
                cal(x,y,a[i].beg,a[i].end);
            else
            {
                f = true;
            }
        }
        if(f)
        {
            printf("Case %d: -1
",cas++);
            continue;
        }
        sort(a,a+n,cmp);
        int cnt = 1;
        double tmp = a[0].end;
        for(int i=1;i<n;i++)
        {
            if(a[i].beg<=tmp)//因为是按结尾排序的，
                //所以a[i].end肯定大于等于tmp，这种情况说明无需添加新的雷达
                continue;
            else//新的端点 起点无法包含，那么重新设置一个雷达（设置在新的线段最右端）
            {
                cnt++;
                tmp = a[i].end;
            }
        }
        printf("Case %d: %d
",cas++,cnt);
    }
    return 0;
}