Check in algorithm topic: Ques20201020

Problem Description

Q1: There is a road AB in a remote mountainous area, and there are n houses along the road AB. The distances from these houses to A are d1, d2,...,dn (d1<d2<...<dn). In order to provide mobile phone services to users in all houses, some base stations need to be set up on this road. In order to ensure communication quality, each house should be located within 4Km from a certain base station. Design an algorithm to find the location of the base station and minimize the total number of base stations.

algorithm

The greedy strategy is to take the distance of d1+4 from the first house to detect whether the next house is covered, if not, then take d2+4 from the next house such as d2. If it has been covered, skip it and check the next one until the check is complete.

Code

//Q1:设有一条边远山区的道路AB，沿着道路AB分布着n所房子。这些房子到A的距离分别是d1,d2,…,dn(d1<d2<…<dn)。
//为了给所有房子的用户提供移动电话服务，需要在这条道路上设置一些基站。为了保证通讯质量，每所房子应该位于距离某个
// 基站的4Km范围内。设计一个算法找基站的位置，并且使得基站的总数最少。

// 算法：贪心策略是，从第一个房子开始取d1+4的距离，检测下一个房子有没有被覆盖，若没有，则在下一个房子如d2开始
// 再取d2+4。若已经覆盖，则跳过，检测下一个，直到检测完为止。

//算法复杂度：

// 关键词：贪心

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue> // 使用队列
#include<cstring>
using namespace std;

class Ques20201020
{
public:

	void test(vector<int>& distances) {
		vector<int> result;
		result.push_back(distances[0]+4); // 从第一个房子开始取d1+4的距离
		for (int i = 1; i < distances.size(); i++) {
			if (result[result.size()-1]+4> distances[i]) {
				continue;
			}
			else {
				result.push_back(distances[i]+4);
			}
		}
		// 打印结果
		for (int i = 0; i < result.size(); i++) {
			cout << " " << result[i];
			cout << endl;
		}

	}

private:
};

//Ques20201020
int main() {
	Ques20201020 ques = Ques20201020();
	vector<int> s = { 2,7,8,12,15 };
	ques.test(s);
	return 0;
}