poj1328Radar Installation(贪心+区间选择)

最新推荐文章于 2022-01-28 16:41:11 发布

luckcircle

最新推荐文章于 2022-01-28 16:41:11 发布

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分类专栏：贪心文章标签： ACM poj1328Radar Install 贪心区间选点

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针对海岸线上雷达安装问题，通过覆盖岛屿最小化雷达数量。利用区间覆盖算法，选择区间最右侧点实现贪婪策略，有效解决实际问题。

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Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 80210		Accepted: 17922

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

Sample Output

Case 1: 2
Case 2: 1
解题思路：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，则可以覆盖该小岛，区间范围的计
算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样，问题即转化为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。每次迭代对于第一个区间，选
择最右边一个点， 因为它可以让较多区间得到满足， 如果不选择第一个区间最右一个点（选择前面的点）， 那么把它换成最右的点之后，以前得到满足的区间， 现在仍然
得到满足， 所以第一个区间的最右一个点为贪婪选择， 选择该点之后， 将得到满足的区间删掉，进行下一步迭代， 直到结束。 
#include 
#include 
#include 
#include 
using namespace std;
struct node{
	double left;
	double right;
}point[1005];
int cmp(node a,node b){
	return a.leftd){
				flag=1;
			}
		}
		if(flag){
			printf("Case %d: -1\n",cas++);
		}
		else{
			sort(point,point+n,cmp);
			temp=point[0].right;
			for(int i=1;itemp){
					sum++;
					temp=point[i].right;
				}
			}
			printf("Case %d: %d\n",cas++,sum);
		}
	}
}