作业题，大致内容是实现普通，专线，双层，郊区4种不同的公交车的信息管理。
我定义了putong，zhuanxian，shuangceng，jiaoqu四个类。

类的声明如下：
#include<iostream>
#include<string>
using namespace std;
static int count=0;
class putong
{
public:
    putong()
    {
        count++;
        num=count;
        grade=1;
    };
    ~putong(){count--;};
    void sethowmany(int sk,int tb)
    {
        sknum=sk;
        tbnum=tb;
    };
    float getsum()
    {
        sum=sknum*0.8+tbnum;
        return sum;
    }
    void upgrade(int c){grade+=c;};
    void setgrade(int g){grade=g;};
    void setname(string n){name=n;};
    string getname(){return name;}
    int getnum(){return num;};
    int getgrade(){return grade;};
protected:
    string name;
    int grade;
    int num;
private:
    float sum;
    int sknum,tbnum;
};

class zhuanxian
{
public:
    zhuanxian()
    {
        count++;
        num=count;
        grade=1;
    };
    ~zhuanxian(){count--;};
    void sethowmany(int sk,int tb)
    {
        sknum=sk;
        tbnum=tb;
    };
    float getsum()
    {
        sum=sknum*1.6+tbnum*2.0;
        return sum;
    }
    void upgrade(int c){grade+=c;};
    void setgrade(int g){grade=g;};
    void setname(string n){name=n;};
    string getname(){return name;}
    int getnum(){return num;};
    int getgrade(){return grade;};
protected:
    string name;
    int grade;
    int num;
private:
    float sum;
    int sknum,tbnum;
};

class shuangceng
{
public:
    shuangceng()
    {
        count++;
        num=count;
        grade=1;
    };
    ~shuangceng(){count--;};
    void sethowmany(int sk,int tb)
    {
        sknum=sk;
        tbnum=tb;
    };
    float getsum()
    {
        sum=sknum*1.4+tbnum*1.5;
        return sum;
    }
    void upgrade(int c){grade+=c;};
    void setgrade(int g){grade=g;};
    void setname(string n){name=n;};
    string getname(){return name;}
    int getnum(){return num;};
    int getgrade(){return grade;};
protected:
    string name;
    int grade;
    int num;
private:
    float sum;
    int sknum,tbnum;
};

class jiaoqu
{
public:
    jiaoqu()
    {
        count++;
        num=count;
        grade=1;
    };
    ~jiaoqu(){count--;};
    void sethowmany(int d){dis=d;};
    float getsum()
    {
        if(dis>=5)
            sum=2;
        else sum=1;
        return sum;
    }
    void upgrade(int c){grade+=c;};
    void setgrade(int g){grade=g;};
    void setname(string n){name=n;};
    string getname(){return name;}
    int getnum(){return num;};
    int getgrade(){return grade;};
protected:
    string name;
    int grade;
    int num;
private:
    float sum;
    int dis;
};





主函数如下：

void main()
{
    putong pt[10];
    zhuanxian zx[10];
    shuangceng sc[10];
    jiaoqu jq[10];


    string n;
    int over;
    do
    {
        static int pts=0,zxs=0,scs=0,jqs=0;
        cout<<"1.查找  2.增加   3.删除"<<endl;
        int o1;
        cin>>o1;
        switch(o1)
        {
            case 1:
            {
                cout<<"1.普通车"<<endl;
                int o2;
                cin>>o2;
                switch(o2)
                {
                    case 1:
                    cout<<"1.编号 2.名称"<<endl;
                    int o3;
                    cin>>o3;
                    switch(o3)
                    {
                        case 1:
                        cout<<"请输入编号："<<endl;
                        int nn;
                        cin>>nn;
                        int i;
                        for(i=0;i<=pts;i++)
                        {    if(pt[i].getnum()==nn)
                            {
                                cout<<"编号："<<pt[i].getnum()<<" 名称："<<pt[i].getname()<<" 级别："<<pt[i].getgrade()<<" 总额:"<<pt[i].getsum()<<endl;
                                break;
                            }
                            else cout<<"查无此车"<<endl;
                            break;
                        }break;
                        
                        case 2:
                        cout<<"请输入名称："<<endl;
                        string nam;
                        cin>>nam;
                        for(i=0;i<=pts;i++)
                        {    if(pt[i].getname()==nam)
                            {
                                cout<<"编号："<<pt[i].getnum()<<" 名称："<<pt[i].getname()<<" 级别："<<pt[i].getgrade()<<" 总额:"<<pt[i].getsum()<<endl;
                                break;
                            }
                            else cout<<"查无此车"<<endl;
                            break;
                        }break;
                    }
                }break;
                
                
            }    

            case 2:
            {
                cout<<"1.普通车 "<<endl;
                int o2;
                cin>>o2;
                switch(o2)
                {
                    case 1:
                    cout<<"输入车名:"<<endl;
                    cin>>n;
                    pt[pts].setname(n);
                    cout<<"输入刷卡总人次："<<endl;
                    int sk;
                    cin>>sk;
                    cout<<"输入投币总人次："<<endl;
                    int tb;
                    cin>>tb;
                    pt[pts].sethowmany(sk,tb);
                    pts++;
                    break;
                }
            }
    
        }
        cout<<"1.继续  2.结束"<<endl;
        cin>>over;
    }while(over==1);
}


我这里只做了和普通车有关的。
现在的问题是，比方说我增加了一条普通车的记录，在最后选择继续，再增加一条。然后再查找的时候，找第二条车的记录的结果是查无此车。我感觉是完成第一条记录返回之后，那个对象就自动被析构掉了？是不是这样？   应该怎么解决呢？