咨寻一个链表嵌套的问题！！求解答！！

我觉得用二叉树 应该更简单点...

#include <iostream>
#include <fstream>
#include <iomanip>
#include <string>
#include <windows.h>
using namespace std;
struct Client
{
string name;
int Phone;
struct Client *Next;
};
struct Business
{
string BusinessName;
Client *ClientHead;
struct Business *Next;
};
int main()
{
Client *ClientMove;
Business *BusinessHead=new Business,*BusinessMove;
BusinessMove=BusinessHead;
int i,num=0;
cout<<"输入要添加的业务个数："<<endl;cin>>i;
while(num<i)
{
num++;
cout<<"请输入第 "<<num<<"个业务名称："<<endl;Business *BusinessData=new Busines;
cin>>BusinessData->BusinessName;
BusinessData->ClientHead=new Client;
BusinessData->ClientHead->Next=nullptr;
BusinessMove->Next=BusinessData;
BusinessMove=BusinessData;
}
BusinessMove->Next=nullptr;
cout<<"业务显示："<<endl;
int num1=1;
BusinessMove=BusinessHead;
while(BusinessMove->Next!=nullptr)
{
BusinessMove=BusinessMove->Next;
cout<<"第 "<<num1<<" 业务是："<<BusinessMove->BusinessName<<endl;
num1++;
}
cout<<endl;
//*****************************
string strname;
cout<<"请输入要添加客户的业务名称："<<endl;cin>>strname;
BusinessMove=BusinessHead;
while(BusinessMove->Next!=nullptr)
{
BusinessMove=BusinessMove->Next;
if(BusinessMove->BusinessName==strname)
{
cout<<"你要添加客户的业务名称是："<<BusinessMove->BusinessName<<endl;
ClientMove=BusinessMove->ClientHead;
break;
}
}
int num2,count=0;
cout<<"请输入要添加客户的个数："<<endl;
cin>>num2;
while(count<num2)
{
count++;
Client *ClientData=new Client;
cout<<"请输入第"<<count<<"客户的姓名，电话:"<<endl;
cin>>ClientData->name>>ClientData->Phone;
ClientMove->Next=ClientData;
ClientMove=ClientData;
}
ClientMove->Next=nullptr;
cout<<"客户输入完毕"<<endl;
//******************
cout<<"输入全部信息："<<endl;
BusinessMove=BusinessHead;
int num3=1,num4=0;
BusinessMove=BusinessHead;
while(BusinessMove->Next!=nullptr)
{
BusinessMove=BusinessMove->Next;
cout<<"第 "<<num3<<" 业务是："<<BusinessMove->BusinessName<<endl;
ClientMove=BusinessMove->ClientHead;
while(ClientMove->Next!=nullptr)
{
num4++;
ClientMove=ClientMove->Next;
cout<<setw(4)<<"第"<<num4<<"个客户信息：";
cout<<setw(14)<<ClientMove->name<<setw(13)<<ClientMove->Phone<<endl;
}
num3++;
}
cout<<endl;
system("pause");
return 0;
}

虽然 楼主神兽会被鄙视，但是我也是闲的蛋疼，就当是练练代码...
基本思路是定义一个Business 和一个Client 结构体；
作为业务链表和客户链表的节点；
当然Business 节点 成员有一个string 用来存储业务名称，和一个 Client 的指针，用来存放Client 链表的头指针，// 最重要就是这点，弄明白了这个基本就不难了.
在初始化Business 的时候都要new一个 Client 指针，用来初始化Client的头指针..
上面写的程序基本功能就是 输入不同业务的名称，以链表方式存储。
然后根据业务名称查询，再初始化客户链表。
当然写的比较仓促，但你的要求达到，当然基本功能不是很完善，楼主要是有空，可以写个菜单，或者添加其他功能，比如增删改查，客户和物业信息之类的...