组合模式

最新推荐文章于 2025-08-16 22:37:45 发布

转载最新推荐文章于 2025-08-16 22:37:45 发布 · 576 阅读

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#设计模式

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本文详细阐述了如何使用C++实现一个企业层级结构的可视化系统，通过抽象类和派生类来构建组织架构，展示从总公司到各个部门的结构，并以递归方式输出各层级信息。

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#include <iostream>
#include <string>
//#include <afx.h>
#include <vector>
#include <list>
using namespace std;

class Company
{
public:
Company(string name) { m_name = name; }
virtual ~Company(){}
virtual void Add(Company *pCom){}
virtual void Show(int depth) {}
protected:
string m_name;
};
//具体公司
class ConcreteCompany : public Company
{
public:
ConcreteCompany(string name): Company(name) {}
virtual ~ConcreteCompany() {}
void Add(Company *pCom) { m_listCompany.push_back(pCom); } //位于树的中间可以增加子树
void Show(int depth)
{
for(int i = 0;i < depth; i++)
  cout<<"-";
cout<<m_name<<"\n";
list<Company *>::iterator iter=m_listCompany.begin();
for(; iter != m_listCompany.end(); iter++) //显示下层结点
  (*iter)->Show(depth + 2);
}
private:
list<Company *> m_listCompany;
};
//具体的部门财务部
class FinanceDepartment : public Company
{
public:
FinanceDepartment(string name):Company(name){}
virtual ~FinanceDepartment() {}
virtual void Show(int depth) //只需显示无限添加函数因为已是叶结点
{
for(int i = 0; i < depth; i++)
cout<<"-";
cout<<m_name<<endl;
}
};
//具体的部门人力资源部
class HRDepartment :public Company
{
public:
HRDepartment(string name):Company(name){}
virtual ~HRDepartment() {}
virtual void Show(int depth) //只需显示无限添加函数因为已是叶结点
{
for(int i = 0; i < depth; i++)
cout<<"-";
cout<<m_name<<endl;
}
};

int main()
{
Company *root = new ConcreteCompany("总公司");
Company *leaf1=new FinanceDepartment("财务部");
Company *leaf2=new HRDepartment("人力资源部");
root->Add(leaf1);
root->Add(leaf2);

//分公司A
Company *mid1 = new ConcreteCompany("分公司A");
Company *leaf3=new FinanceDepartment("财务部");
Company *leaf4=new HRDepartment("人力资源部");
mid1->Add(leaf3);
mid1->Add(leaf4);
root->Add(mid1);
//分公司B
Company *mid2=new ConcreteCompany("分公司B");
FinanceDepartment *leaf5=new FinanceDepartment("财务部");
HRDepartment *leaf6=new HRDepartment("人力资源部");
mid2->Add(leaf5);
mid2->Add(leaf6);
root->Add(mid2);
root->Show(0);

delete leaf1; delete leaf2;
delete leaf3; delete leaf4;
delete leaf5; delete leaf6;
delete mid1; delete mid2;
delete root;
return 0;
}