Check in algorithm topic: Ques20201012

1. Event scheduling issues

1. Problem description

 Event scheduling problem, one venue, arrange as many activities as possible.

2. Code

// 活动安排

// 算法描述：①按照结束时间从大到小排序-->②优先安排结束时间早的-->③判断下一个活动的相容性的方法是f[i]>s[j]

// 关键词：贪心


#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct Activity
{
	int startTime;
	int finishTime;
	int label;
};
class Ques20201012
{
public:
	/*vector<Activity> activities;
	
	Ques20201012(vector<int> start, vector<int> finish) {
		Activity ac = Activity();
		for (int i = 0; i <= start.size();i++) {
			ac.label = i;
			ac.startTime = start[i];
			ac.finishTime = finish[i];
			activities.push_back(ac);
		}
		
	};*/
	void test(vector<Activity> &activities) {
		// 打印
		
		vector<int> result=greedyActivity(activities);
		cout << "最终安排结果是：" << endl;
		for (int i = 0; i < result.size();i++) {
			
			cout << result[i]<<" , ";
		}
	}


private:
	
	vector<int> greedyActivity(vector<Activity>& activities) {
		int j = 0;
		vector<int> solution = { 0 };
		for (int i = 1; i < activities.size();i++) {
			if (activities[i].startTime>= activities[j].finishTime) {
				solution.push_back(activities[i].label);
				j = i;
			}
		}
		return solution;
	}

};

bool cmp(const Activity& a, const Activity& b)//自定义结构体的排序,sort排序只能在main函数在调用
{
	return a.finishTime < b.finishTime;
}

int main() {
	vector<int> s = {1,3,0,5,3,5,6,8,8,2,12};
	vector<int> f = { 4,5,6,7,8,9,10,11,12,13,14 };
	vector<Activity> activities;
	for (int i = 0; i < 11; i++) {
		Activity ac = Activity();
		ac.label = i;
		ac.startTime = s[i];
		ac.finishTime = f[i];
		activities.push_back(ac);
	}
	Ques20201012 qus = Ques20201012();
	sort(activities.begin(), activities.end(), cmp); // 自定义结构体排序，vector排序

	qus.test(activities);
	return 0;
}

3. Proof of Greedy Strategy

Second, the backpack problem

Similar to the previous question, but the greedy strategy is based on unit value, and the last one is directly installed.

1. Code

class Ques20201012Q2
{
public:
	

	void test(vector<Knapsack>& goods) {
		// 打印
		vector<int> result = greedyKnapsack(goods,50);
		cout << "最终安排结果是：" << endl;
		for (int i = 0; i < result.size(); i++) {

			cout << result[i] << " , ";
		}
	}


private:

	vector<int> greedyKnapsack(vector<Knapsack>& napsacks, float M) {//M是总重量
		vector<int> solution;// 解向量
		float remain = M;
		int i;
		for (i= 0; i < napsacks.size();i++) {
			if (napsacks[i].weight<= remain) {
				solution.push_back(napsacks[i].label);
				remain = remain - napsacks[i].weight;
			}
			else
			{
				break;
			}
		}
		if (i<= napsacks.size()) {
			float lastOne = remain / napsacks[i].weight;
			if (lastOne!=0) {
				solution.push_back(napsacks[i].label);
				cout << "最后一项装的是重量为:"<<lastOne<<endl;
			}
			
		}
		return solution;
	}
};

bool cmp2(const Knapsack& a, const Knapsack& b)//自定义结构体的排序,sort排序只能在main函数在调用
{
	return a.unitValue > b.unitValue; // 这是增序
}
int main() {

	// Q2 背包问题

	vector<int> w = { 10,20,30 };
	vector<int> v = { 60,100,120 };
	vector<Knapsack> napsacks;
	for (int i = 0; i < w.size(); i++) {
		Knapsack na = Knapsack(w[i], v[i]);
		na.label = i;
		napsacks.push_back(na);
	}
	Ques20201012Q2 kna = Ques20201012Q2();
	sort(napsacks.begin(), napsacks.end(), cmp2); // 自定义结构体排序，vector排序

	kna.test(napsacks);




	return 0;
}

2. Greed proof