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BIGHOMEWORKS/1/第一次作业参考答案.md

@@ -1,19 +1,19 @@
-### ��һ���
+### 第一题答案
 ```c++
 #include "LinKList.h"
-bool Intersect_Arr(SqList &La,SqList Lb) //��˳���ʵ��
+bool Intersect_Arr(SqList &La,SqList Lb) //用顺序表实现
 {
    ElemType temp;
-    while(!ListEmpty_Sq(La))     //La����Ԫ����δ��δ������
+    while(!ListEmpty_Sq(La))     //La表的元素尚未处未处理完
     {
-        GetElem_Sq(La,0,temp);    //��La��ɾ����һ������Ԫ�ظ�ֵ��temp
-        if(LocateElem_Sq(Lb,temp)==-1)     //��Lb�в�����ֵ��temp��ȵ�����Ԫ��
-              ListDelete_Sq(La,0,e);     //����������La�����һ������Ԫ��֮��
+        GetElem_Sq(La,0,temp);    //从La中删除第一个数据元素赋值给temp
+        if(LocateElem_Sq(Lb,temp)==-1)     //若Lb中不存在值和temp相等的数据元素
+              ListDelete_Sq(La,0,e);     //则将它插入在La中最后一个数据元素之后
     }
-    DestroyList_Sq(Lb);               //�������Ա� 
+    DestroyList_Sq(Lb);               //撤销线性表 
 }
 
-bool Intersect_L(LinkList &La,LinkList Lb){   //������ʵ��
+bool Intersect_L(LinkList &La,LinkList Lb){   //用链表实现
     ElemType temp;
     LinkList p,q;
     p=La->next;
@@ -32,8 +32,8 @@ bool Intersect_L(LinkList &La,LinkList Lb){   //
 ```
 
 
-### �ڶ���𰸣�
-- ������㷨˼���ǣ����ҵ����������ı�βָ�룬����һ�������ı�βָ����ڶ���������ͷ���������������ʹ֮��Ϊѭ���ġ�ʵ�ֱ��⹦�ܵĺ�������:
+### 第二题答案：
+- 本题的算法思想是：先找到两个链表的表尾指针，将第一个链表的表尾指针域第二个链表的头结点链接起来，再使之成为循环的。实现本题功能的函数如下:
 ```c++
 #include "LinkList.h"
 
@@ -47,11 +47,11 @@ void link(LinkList La, LinkList Lb)
 }
 ```
 
-### ������𰸣�
-- ����������ͷ�������ͬ���ַ�����Ϊ�����ġ������硰abba���͡�abcba���ǻ��ģ�����abcde���͡�ababab�����ǻ��ġ��ٱȽ����Ա�����ջ�Ͷ��еȷ�ʽ��
-- 1.�������ַ�ʽʵ�ֻ��ĵ��жϺ��ʣ�
-- 2.��д���㷨�б�����һ���ԡ�@��Ϊ���������ַ������Ƿ��ǻ��ġ�
-- ����Ȼ���ö�ջ�Ͷ��н�ϵķ�ʽʵ�ֺ���
+### 第三题答案：
+- 假设称正读和反读都相同的字符序列为“回文”，例如“abba”和“abcba”是回文，而“abcde”和“ababab”则不是回文。①比较线性表、堆栈和队列等方式，
+- 1.解释哪种方式实现回文的判断合适？
+- 2.试写出算法判别读入的一个以“@”为结束符的字符序列是否是回文。
+- 答：显然是用堆栈和队列结合的方式实现合适
 ```c++
 #include "SqQueue.h"
 #include "SqStack.h"
@@ -72,27 +72,27 @@ bool IsHuiWen(char *str)
   {
     if(DeQueue_Sq(Q,x)&&Pop_Sq(S,y)&&x!=y)
 	{
-      cout<<str<<"���ǻ��ģ�"<<endl;
+      cout<<str<<"不是回文！"<<endl;
       return false;
     }
   }
   if(!QueueEmpty_Sq(Q)||!StackEmpty_Sq(S))
     {
-      cout<<str<<"���ǻ��ģ�"<<endl;
+      cout<<str<<"不是回文！"<<endl;
       return false;
     }
   else
     {
-      cout<<str<<"�ǻ��ģ�"<<endl;
+      cout<<str<<"是回文！"<<endl;
       return true;
     }
 }
 ```
 
-������𰸣�
-- ��֪Q��һ��������ͷ���ģ��ǿն��У�S��һ��������ͷ���ģ���ջ
-- ��д�㷨ʵ�֣�������Q�е�����Ԫ�����á�Ҫ�󣺵���ͷ�ļ���������ʵ�֡�
-- ��Ԫ�����γ����У�������ջ�������ջ�������˶���
+第四题答案：
+- 已知Q是一个（不带头结点的）非空队列，S是一个（不带头结点的）空栈
+- 编写算法实现：将队列Q中的所有元素逆置。要求：调用头文件操作函数实现。
+- 答：元素依次出队列，依次入栈，逆序出栈，依次人队列
 ```c++
 #include "SqQueue.h"
 #include "SqStack.h"
@@ -110,4 +110,236 @@ void Inverse(SqQueue &Q, SqQueue &S){
     }
 }
 ```
-������𰸣�
+第五题答案：
+```
+typedef struct
+{
+    ElemType *queue;
+    int front;
+    int rear;
+    int Count;
+    int queuesize;
+    int incrementsize;
+} SqQueue;
+
+//初始化操作
+void InitQueue_Sq(SqQueue &Q,int maxsize=QUENU_INIT_SIZE,int incresize=QUENUINCREMENT)
+{
+    Q.queue=(ElemType*)malloc(maxsize*sizeof(ElemType));
+    if(!Q.queue)
+        exit(1);
+    Q.front=Q.rear=Q.Count=0;
+    Q.queuesize=maxsize;
+    Q.incrementsize=incresize;
+}
+
+//求队列的长度
+int QueueLength_Sq(SqQueue Q)
+{
+    return (Q.rear-Q.front+Q.queuesize)%Q.queuesize;
+}
+
+//进队操作
+bool EnQueue_Sq(SqQueue &Q,ElemType e)
+{
+    //cout<<"未满";
+    //从队尾插入元素，成功则返回true
+    if(Q.Count==Q.queuesize) //队满给顺序队列增补空间
+    {
+        cout<<"队满";
+        Q.queue=(ElemType*)realloc(Q.queue,(Q.queuesize+Q.incrementsize)*sizeof(ElemType));
+        if(!Q.queue)
+            return false;
+        if(Q.front>=Q.rear)   //队首指针在队尾指针后面，重新确定位置
+        {
+            for(int i=Q.queuesize; i>=Q.front; i--)
+            {
+                Q.queue[i+Q.incrementsize]=Q.queue[i];
+            }
+            Q.front+=Q.incrementsize;
+        }
+        Q.queuesize+=Q.incrementsize;
+    }
+    Q.queue[Q.rear]=e;
+    Q.rear=(Q.rear+1)%Q.queuesize;
+    Q.Count++;
+    return true;
+}
+
+//出队操作
+bool DeQueue_Sq(SqQueue &Q,ElemType &e)
+{
+    if(Q.Count==0)
+    {
+        return false;
+    }
+    e=Q.queue[Q.front];
+    Q.front=(Q.front+1)%Q.queuesize;
+    Q.Count--;
+    return true;
+}
+```
+第六题：
+```
+typedef struct{
+    short x;
+    short y;
+}ElemType;
+#include "SqStack.h"
+
+void TraveseMap(int map[][10]){
+    int i,j;
+    for(i=0;i<10;i++){
+        for(j=0;j<10;j++){
+            if(map[i][j]==0){
+                cout<<"##";
+            }else if(map[i][j]==1){
+                cout<<"  ";
+            }else if(map[i][j]==3){
+                cout<<"  ";
+            }else if(map[i][j]==4){
+                cout<<"〇";
+            }
+        }
+        cout<<endl;
+    }
+
+}
+
+int look_around(int map[][10],ElemType &e){
+    //·µ»Ø0±íʾÎÞ·¿É×ߣ¬1±íʾÕÒµ½Â·
+    if(map[e.x-1][e.y]==1){
+        e.x=e.x-1;
+        return 1;
+    }else if(map[e.x+1][e.y]==1){
+        e.x=e.x+1;
+        return 1;
+    }else if(map[e.x][e.y-1]==1){
+        e.y=e.y-1;
+        return 1;
+    }else if(map[e.x][e.y+1]==1){
+        e.y=e.y+1;
+        return 1;
+    }else {
+        return 0;
+    }
+
+}
+
+void findroad(int map[][10],ElemType start,ElemType end){
+    SqStack mystack;
+    InitStack_Sq(mystack);
+    ElemType temp=start;
+    map[start.x][start.y]=3;
+    Push_Sq(mystack,temp);
+    int counts=1;
+    while(true){
+        //cout<<counts++<<endl;
+        int result=look_around(map,temp);
+        if(result==1){
+            if(temp.x==end.x&&temp.y==end.y){
+                Push_Sq(mystack,end);
+                break;
+            }
+            map[temp.x][temp.y]=3;
+            Push_Sq(mystack,temp);
+        }else if(result==0){
+            Pop_Sq(mystack,temp);
+            GetTop_Sq(mystack,temp);
+        }
+        //TraveseMap(map);
+    }
+
+    while(!StackEmpty_Sq(mystack)){
+        Pop_Sq(mystack,temp);
+        map[temp.x][temp.y]=4;
+    }
+}
+
+
+int main()
+{
+    int Map[10][10] =
+{  {0,0,0,0,0,0,0,0,0,0}, //0ÐÐ
+   {0,1,1,0,1,1,1,0,1,0}, //1ÐÐ
+   {0,1,1,0,1,1,1,0,1,0}, //2ÐÐ
+   {0,1,1,1,1,0,0,1,1,0}, //3ÐÐ
+   {0,1,0,0,0,1,1,1,1,0}, //4ÐÐ
+   {0,1,1,1,0,1,1,1,1,0}, //5ÐÐ
+   {0,1,0,1,1,1,0,1,1,0}, //6ÐÐ
+   {0,1,0,0,0,1,0,0,1,0}, //7ÐÐ
+   {0,0,1,1,1,1,1,1,1,0}, //8ÐÐ
+   {0,0,0,0,0,0,0,0,0,0}  //9 ÐÐ
+};
+    ElemType start={1,1};
+    ElemType end={8,8};
+    findroad(Map,start,end);
+    TraveseMap(Map);
+    return 0;
+}
+```
+第七题
+```
+void wait_in_line(SqQueue &Q){
+    cout<<"请输入病历号";
+    ElemType i;
+    cin>>i;
+    EnQueue_Sq(Q,i);
+}
+
+void see_a_doctor(SqQueue &Q){
+    ElemType e;
+    DeQueue_Sq(Q,e);
+    cout<<e<<"号病人就诊"<<endl;;
+}
+
+void see_queue(SqQueue &Q){
+    for(int i=Q.front;i!=Q.rear;i=(i+Q.queuesize+1)%Q.queuesize){
+        cout<<setw(5)<<Q.queue[i];
+    }
+    cout<<endl;
+}
+
+void exit_line(SqQueue &Q){
+    while(!QueueEmpty_Sq(Q)){
+        ElemType e;
+        DeQueue_Sq(Q,e);
+        cout<<setw(5)<<e;
+    }
+    cout<<endl;
+    exit(0);
+}
+
+void get_off_work(){
+    exit(0);
+}
+
+int main()
+{
+    SqQueue myqueue;
+    InitQueue_Sq(myqueue);
+    bool working=true;
+    while(working){
+        cout<<"\n**********************************************************"<<endl;
+        cout<<"*    欢迎来到黑诊所\n";
+        cout<<"*    1、排队"<<endl;
+        cout<<"*    2、就诊"<<endl;
+        cout<<"*    3、查看排队"<<endl;
+        cout<<"*    4、不再排队"<<endl;
+        cout<<"*    5、下班"<<endl;
+        cout<<"*    请选择操作"<<endl;
+        int iOperator;
+        cin>>iOperator;
+        switch(iOperator){
+            case 1:wait_in_line(myqueue);break;
+            case 2:see_a_doctor(myqueue);break;
+            case 3:see_queue(myqueue);break;
+            case 4:exit_line(myqueue);break;
+            case 5:get_off_work();break;
+            default:cout<<"错误";break;
+        }
+    }
+    return 0;
+}
+```
+