poj 1328 Radar Installation 已知一定数量的区间,求最小数量的点,使得每个区间内斗至少存在一个点 贪心

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23954		Accepted: 5130

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

#include
#include
#include
#include
#include
using namespace std;
//已知一定数量的区间,求最小数量的点,使得每个区间内斗至少存在一个点
//贪心
struct Node
{
    int x,y;
    double left,right;//雷达能覆盖小岛的最左最右坐标
};
Node a[1100];
int n,d;
bool cmp(Node h,Node k)
{
    if(fabs(h.left-k.left)<1e-6) return h.right>k.right;
    return h.leftd) flag=0;
            double tmp=sqrt(d*d-a[i].y*a[i].y+0.0);
            a[i].left=a[i].x-tmp,a[i].right=a[i].x+tmp;
        }
        if(!flag){printf("Case %d: -1/n",pl++);continue;}
        sort(a,a+n,cmp);
        int cnt=1;double currentRight=a[0].right;
        for(int i=1;icurrentRight) cnt++,currentRight=a[i].right;
            else ;
        }
        printf("Case %d: %d/n",pl++,cnt);
    }
    return 0;
}