POJ 1328 Radar Installation（贪心）

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
首先考虑把坐标维度降下来再考虑贪心算法
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<cmath>
typedef long long LL;
using namespace std;
double d,x,y;
struct node{
    double l,r;
}e[1010];
int cmp(node l1,node l2)
{
    return l1.r<l2.r;
}
int main()
{
    int n,flag;
    int cas=1;
    while(~scanf("%d%lf",&n,&d))
    {
        if(n==0&&d==0)   break;
        flag=1;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&x,&y);
            if(!flag)  continue;
            if(d<y)  flag=0;
            double dis=sqrt(d*d-y*y);
            e[i].l=x-dis;
            e[i].r=x+dis;
        }
        printf("Case %d: ",cas++);
        if(!flag)
        {
            printf("-1
");
            continue;
        }
        sort(e,e+n,cmp);
        int ans=1;
        double pos=e[0].r;
        for(int i=1;i<n;i++)
        {
            if(e[i].l>pos)
            {
                ans++;
                pos=e[i].r;
            }
            if(e[i].r<pos)
                pos=e[i].r;
        }
        printf("%d
",ans);
    }
    return 0;
}