SDUST数据结构

选择题:

 

 

 

 

 

 

 

 

 

 

函数题:

  6-1 二分查找:

裁判测试程序样例:

#include <stdio.h>
#include <stdlib.h>

#define MAXSIZE 10
#define NotFound 0
typedef int ElementType;

typedef int Position;
typedef struct LNode *List;
struct LNode {
    ElementType Data[MAXSIZE];
    Position Last; /* 保存线性表中最后一个元素的位置 */
};

List ReadInput(); /* 裁判实现,细节不表。元素从下标1开始存储 */
Position BinarySearch( List L, ElementType X );

int main()
{
    List L;
    ElementType X;
    Position P;

    L = ReadInput();
    scanf("%d", &X);
    P = BinarySearch( L, X );
    printf("%d
", P);

    return 0;
}

/* 你的代码将被嵌在这里 */

输入样例1:

5
12 31 55 89 101
31

输出样例1:

2

输入样例2:

3
26 78 233
31

输出样例2:

0

代码:

Position BinarySearch( List L, ElementType X )
{
    if(L == NULL)
        return NotFound;//不符合条件,空 
    int low=1, high=L->Last;//设置左右指针 
    int mid;
    while(low<=high)//左右指针相遇跳出循环 
    {
        mid = (low+high)/2;
        if(X>L->Data[mid])    
            low = mid+1;
        else if(X<L->Data[mid])
            high = mid-1;
        else
            return mid;
    }    
    return NotFound;//未找到,空 
} 
View Code

  6-2 线性探测法的查找函数:

裁判测试程序样例:

#include <stdio.h>

#define MAXTABLESIZE 100000  /* 允许开辟的最大散列表长度 */
typedef int ElementType;     /* 关键词类型用整型 */
typedef int Index;           /* 散列地址类型 */
typedef Index Position;      /* 数据所在位置与散列地址是同一类型 */
/* 散列单元状态类型,分别对应:有合法元素、空单元、有已删除元素 */
typedef enum { Legitimate, Empty, Deleted } EntryType;

typedef struct HashEntry Cell; /* 散列表单元类型 */
struct HashEntry{
    ElementType Data; /* 存放元素 */
    EntryType Info;   /* 单元状态 */
};

typedef struct TblNode *HashTable; /* 散列表类型 */
struct TblNode {   /* 散列表结点定义 */
    int TableSize; /* 表的最大长度 */
    Cell *Cells;   /* 存放散列单元数据的数组 */
};

HashTable BuildTable(); /* 裁判实现,细节不表 */
Position Hash( ElementType Key, int TableSize )
{
    return (Key % TableSize);
}

#define ERROR -1
Position Find( HashTable H, ElementType Key );

int main()
{
    HashTable H;
    ElementType Key;
    Position P;

    H = BuildTable(); 
    scanf("%d", &Key);
    P = Find(H, Key);
    if (P==ERROR)
        printf("ERROR: %d is not found and the table is full.
", Key);
    else if (H->Cells[P].Info == Legitimate)
        printf("%d is at position %d.
", Key, P);
    else
        printf("%d is not found.  Position %d is returned.
", Key, P);

    return 0;
}

/* 你的代码将被嵌在这里 */

输入样例1:(-1表示该位置为空)

11
11 88 21 -1 -1 5 16 7 6 38 10
38

输出样例1:

38 is at position 9.

输入样例2:

11
11 88 21 -1 -1 5 16 7 6 38 10
41

输出样例2:

41 is not found.  Position 3 is returned.

输入样例3:

11
11 88 21 3 14 5 16 7 6 38 10
41

输出样例3:

ERROR: 41 is not found and the table is full.

代码:

Position Find( HashTable H, ElementType Key )
{
    Position p, p1;
    int cts=0;
    p = p1 =Hash(Key, H->TableSize);
    while(H->Cells[p].Data!=Key&&H->Cells[p].Info!=Empty)
    {
        cts++;
        if(cts == MAXTABLESIZE)
            return ERROR;//若没有空单元,则返回ERROR
        p = (p1 + cts)%H->TableSize;
    }
    return p;
}
View Code

  6-3 分离链接法的删除操作函数:

裁判测试程序样例:

#include <stdio.h>
#include <string.h>

#define KEYLENGTH 15                   /* 关键词字符串的最大长度 */
typedef char ElementType[KEYLENGTH+1]; /* 关键词类型用字符串 */
typedef int Index;                     /* 散列地址类型 */
typedef enum {false, true} bool;

typedef struct LNode *PtrToLNode;
struct LNode {
    ElementType Data;
    PtrToLNode Next;
};
typedef PtrToLNode Position;
typedef PtrToLNode List;

typedef struct TblNode *HashTable; /* 散列表类型 */
struct TblNode {   /* 散列表结点定义 */
    int TableSize; /* 表的最大长度 */
    List Heads;    /* 指向链表头结点的数组 */
};

Index Hash( ElementType Key, int TableSize )
{
    return (Key[0]-'a')%TableSize;
}

HashTable BuildTable(); /* 裁判实现,细节不表 */
bool Delete( HashTable H, ElementType Key );

int main()
{
    HashTable H;
    ElementType Key;

    H = BuildTable(); 
    scanf("%s", Key);
    if (Delete(H, Key) == false)
        printf("ERROR: %s is not found
", Key);
    if (Delete(H, Key) == true)
        printf("Are you kidding me?
");
    return 0;
}

/* 你的代码将被嵌在这里 */

输入样例1:

able

输出样例1:

able is deleted from list Heads[0]

输入样例2:(散列表如样例1图)

date

输出样例2:

ERROR: date is not found

代码:

bool Delete( HashTable H, ElementType Key )
{
    Position p,q;
    Index pos;//关键字地址
    pos = Hash(Key, H->TableSize);//初始化散列位置
    p = H->Heads[pos].Next;//从该邻接表第一个节点开始
    while(p&&strcmp(p->Data, Key))//当未到表尾,并且Key未找到时
    {
        q = p;
        p = p->Next;
    } 
    if(p==NULL)
        return false;
    else
    {
        printf("%s is deleted from list Heads[%d]",Key,pos);
        q = p->Next;
        free(p);//删除p节点,释放p内存
        return true;
    }
}
View Code

编程题:

  7-1 两个有序序列的中位数:

输入样例1:

5
1 3 5 7 9
2 3 4 5 6

输出样例1:

4

输入样例2:

6
-100 -10 1 1 1 1
-50 0 2 3 4 5

输出样例2:

1

代码:

#include<stdio.h>
int main()
{
    int n;
    scanf("%d",&n);
    int a[100001]={0},b[100001]={0},c[200002]={0};//a,b分别用于存储两行数,c用于排序的时候按顺序填入 
    for(int i=0;i<2;i++)//读入两个数组的元素 
    {
        if(i==0)
        {
            for(int j=0;j<n;j++)
                scanf("%d",&a[j]);    
        }
        if(i==1)
        {
            for(int j=0;j<n;j++)
                scanf("%d",&b[j]);    
        }
    }
    int x=0,y=0;
    int i=0;
    while(x<n&&y<n)//分别进行比较, 按照升序依次填入c数组中, 
    {
        if(a[x] > b[y])
        {
            c[i] = b[y];
            i++;
            y++; 
        } 
        else
        {
            c[i] = a[x];
            i++;
            x++;
        }
    }
    while(y<n)//
    {
        c[i++] = a[y++];
    }
    while(x<n)//
    {
        c[i++] = b[x++];
    }
    int mid = (i-1)/2;//寻找中位数, 
    printf("%d",c[mid]);
    return 0;
}
View Code

  7-2 是否同一棵二叉搜索树:

输入样例:

4 2
3 1 4 2
3 4 1 2
3 2 4 1
2 1
2 1
1 2
0

输出样例:

Yes
No
No

代码:

#include<iostream>
using namespace std;
typedef struct Tree
{
    int data;
    struct Tree *Left;
    struct Tree *Right;    
}TNode, *BTree;
void BuildT(BTree &T, int a)
{
    if(T == NULL)
    {
        T = new Tree;
        T->data = a;
        T->Left = T->Right = NULL;
    }
    else
    {
        if(a<T->data)
            BuildT(T->Left, a);
        else
            BuildT(T->Right, a); 
    }
}
int panduan(Tree *t1, Tree *t2)
{
    if(!t1&&!t2)
    return 1;
    if(t1&&t2)
        if(t1->data==t2->data )
        if(panduan(t1->Left,t2->Left )&&panduan(t1->Right,t2->Right ))
        return 1;
    return 0;
}
int main()
{
    int n, l,a;
    int flag=0;
    while(cin>>n)
    {
        Tree *t = NULL;
        if(n == 0)
            return 0;
        cin>>l;
        for(int i=0;i<n;i++)
        {
            cin>>a;
            BuildT(t, a);
        }
        while(l--)
        {
            Tree *tt = NULL;
            for(int i=0;i<n;i++)
            {
                cin>>a;
                BuildT(tt, a);
            }
            flag = panduan(t, tt);
            if(flag==0)
                cout<<"No"<<endl;
            else
                cout<<"Yes"<<endl;
        }
    }
    return 0;
}
View Code

  7-3 是否完全二叉搜索树:

输入样例1:

9
38 45 42 24 58 30 67 12 51

输出样例1:

38 45 24 58 42 30 12 67 51
YES

输入样例2:

8
38 24 12 45 58 67 42 51

输出样例2:

38 45 24 58 42 12 67 51
NO

代码:

#include<stdio.h>
#include <string.h>
int a[105];
int n,m;
void insert(int p, int data)
{
    if(a[p] == 0)
    {
        a[p] = data;
        return ;
    }
    if(data > a[p])
        insert(p*2, data);
    else
        insert(p*2+1, data);
}
int main()
{
    //int n,m;
    //int a[105];
    scanf("%d",&n);
    memset(a, 0, sizeof(int));
    for(int i=0;i<n;i++)
    {
        scanf("%d",&m);
        insert(1, m);    
    }    
    int nn=0;;
    int i=0;;
    while(nn<n)
    {
        while(a[i] == 0)
            i++;
        if(nn)
            printf(" %d", a[i]);
        else
            printf("%d",a[i]);
        i++;
        nn++;
    }
    if(i == n+1)
        printf("
YES
");
    else
        printf("
NO
");
}
View Code
原文地址:https://www.cnblogs.com/3cH0-Nu1L/p/14055394.html