《C语言》数据机构实验报告.docx

1、C语言数据机构实验报告数据结构实验报告实验一：抽象数据类型的实现-以集合为例问题：以集合为例设计一个抽象数据类型；1. 集合是对象的无序组合，我们可以把集合定义成一个新的数据类型;2. 为了简化可以把集合限制为有限整数元素的集合，有了这个限制，集合可以作为一个静态整数数组实现；3. 为了 操作方便，在处理对象中还包括集合的元素个数（即：集合基数）。C语言程序代码：/-#include #include#define maxCard 16#define maxLen 64#define noErr 1#define overflow 0/-/定义结构体Set和Set_productstruct

2、Set int elemsmaxCard; int card; ;struct Set_product char elemsmaxCard*maxCardmaxLen; int card; ;/-/ 函数声明void EmptySet(Set *);int AddElem(Set *, int );void Print(Set *);void RmvElem(Set *,int );bool Member(Set *,int );void Copy(Set * , Set * );void Intersect(Set * , Set * , Set * );int Union(Set * ,

3、Set * , Set * );bool Equal(Set * , Set * );bool Subset(Set * , Set * );bool PSubset(Set * , Set * );void Cartesian_product(Set * , Set * , Set_product *);void Print(Set_product *);/-/ 主函数调用int main() Set s1,s2,s3; Set_product ss1; EmptySet(&s1); EmptySet(&s2); EmptySet(&s3); AddElem(&s1,10); AddElem

4、(&s1,20); AddElem(&s1,30); AddElem(&s1,40); AddElem(&s2,30); AddElem(&s2,50); AddElem(&s2,10); AddElem(&s2,60); couts1 = ; Print(&s1); couts2 = ; Print(&s2); RmvElem(&s2,50); couts2 - 50 = ; Print(&s2); if (Member(&s1,20) cout20 is in s1 endl; Intersect(&s1, &s2, &s3); couts1 intsec s2 = ; Print(&s3

5、); Union(&s1, &s2, &s3); couts1 union s2 = ; Print(&s3); if (Subset(&s1,&s3) couts1 is the subset of s3 endl; if (PSubset(&s1,&s3) couts1 is the pure subset of s3 endl; if (!Equal(&s1,&s2) couts1 /= s2 endl; else couts1 = s2 endl; Cartesian_product(&s1, &s2, &ss1); coutcard = 0; int AddElem(Set *set

6、, int elem) for (int i=0; icard; +i) if (set-elemsi=elem) return noErr; if (set-cardelemsset-card+=elem; return noErr; else return overflow; /AddElem/-/函数Print void Print(Set *set) if (set-card=0)/ cout The Set has no element!n; cout endl; else cout ; for (int i =0; icard-1; +i) cout elemsicard0) co

7、ut elemsset-card-1; cout card=0)/ cout The Set has no element!n; cout endl; else cout ; for (int i =0; icard-1; +i) cout elemsicard0) cout elemsset-card-1; cout endl; /Print Cartesian_product/-/函数RmvElemvoid RmvElem(Set *set,int elem) for (int i=0; icard; +i) if (set-elemsi=elem) for (; icard-1;+i)

8、set-elemsi=set-elemsi+1; -set-card; /RmvElem/-/函数Memberbool Member(Set *set,int elem) for(int i=0; icard; +i) if (set-elemsi=elem) return true; return false; /Member/-/函数Copy void Copy(Set * set, Set * res) for (int i =0; icard; +i) res-elemsi=set-elemsi; res-card=set-card; /Copy/-/函数Equal bool Equa

9、l(Set * set1, Set * set2) if (set1-card!=set2-card) return false; for (int i=0; icard; +i) if (!Member(set2,set1-elemsi) return false; return true; /Equal /-/函数Intersectvoid Intersect(Set * set1, Set * set2, Set * res) res-card=0; for (int i=0;icard; +i) for (int j=0;jcard;+j) if (set1-elemsi=set2-e

10、lemsj) res-elemsres-card+=set1-elemsi; break; /Intersect /-/函数Union int Union(Set * set1, Set * set2, Set * res) Copy(set1, res); for (int i=0;icard;+i) if (AddElem(res, set2-elemsi)=overflow) return overflow; return noErr; /Union/-/函数Subsetbool Subset(Set * set1, Set * set2) if (set1-card set2-card

11、) return false; for (int i=0;icard;+i) if (!Member(set2,set1-elemsi) return false; return true; /Subset/-/函数PSubsetbool PSubset(Set * set1, Set * set2) if (set1-card card & Subset(set1,set2) return true; /PSubset/-/函数Cartesian_productvoid Cartesian_product(Set * set1, Set * set2, Set_product *res) i

12、nt k=0; res-card=set1-card*set2-card; for (int i=0 ;icard; +i) for (int j=0;jcard;+j,+k) sprintf ( res-elemsk, (%d,%d), set1-elemsi, set2-elemsj); 实验二： 链表及其应用问题：试编一个算法，将一个用单链表表示的稀疏多项式分解两个多项式。1.这两个多项式中各自仅含奇次项或偶次项；2.要求利用原链表中的节点空间构成这两个多项式。C语言程序代码：#include #include typedef struct float coef; int expnum;

13、 term, ElemType; /数据包括底数，次数；typedef struct LNode ElemType data; struct LNode *next;LNode, *LinkList; /声明节点为Lnode型和LinkList的指针；void CreateList_L(LinkList&L, int n );void Display_L(LinkList L);void Split_L(LinkList &plyn, LinkList &odd);void Reverse_L(LinkList L);void main() int NodeNum = 4; LinkList

14、List, oddList; CreateList_L(List, NodeNum); Display_L(List); Reverse_L(List); Display_L(List); Split_L(List, oddList); Display_L(List); Display_L(oddList); / 输入n个元素的值，利用尾插法建立带表头结点的单链线性表L void CreateList_L(LinkList&L, int n) LinkList p,Pre; int i; L = (LinkList)malloc(sizeof(LNode);/先建立一个带头节点的线性链表； P

15、re=L; for (i=0; idata.expnum); printf(the coef is :); scanf(%f,&p-data.coef); /输入数据的底数； Pre-next = p; Pre = p; / 修改尾指针； Pre-next=NULL; void Display_L(LinkList L)/输出次数与底数； L = L-next; while (L) printf(The expnum is :);/次数； printf(%dn,L-data.expnum); printf(The coef is :); printf(%.2fn,L-data.coef); L

16、 = L-next; printf(n);void Reverse_L(LinkList L) LinkList p = NULL, q, s; q = L-next; while (q) s = q-next; q-next = p; p = q; q = s; L-next = p;/Reverse_Lvoid Split_L(LinkList &plyn, LinkList &odd) LinkList p,q,s; odd = (LinkList)malloc(sizeof(LNode); odd-next= NULL; q=plyn; p=plyn-next; s=odd; whil

17、e(p) if (p-data.expnum%2 = 0) q=p; p=p-next; else q-next=p-next; /在原表中删除奇次项节点 s-next=p; /将奇次项节点插入到新表中 p=q-next; s=s-next; s-next=NULL; /Split_L实验三：队列和栈的使用问题：试编写一个算法，利用栈将队列逆置；1.栈和队列的数据元素都使用整型变量；2.队列用链式存储，栈用动态的顺序存储；C语言程序代码：/-#include /#include #include /-typedef struct float coef; int expnum; term, El

18、emType; typedef struct LNode ElemType data; struct LNode *next;LNode, *LinkList;/-void CreateList_L(LinkList &, int );void Display_L(LinkList );void Split_L(LinkList &, LinkList &);void Reverse_L(LinkList );/-void main() int NodeNum = 4; LinkList List, oddList; CreateList_L(List, NodeNum); Display_L

19、(List); Reverse_L(List); Display_L(List); Split_L(List, oddList); Display_L(List); Display_L(oddList); /main/-void CreateList_L(LinkList &L, int n) / 逆位序输入n个元素的值，建立带表头结点的单链线性表L LinkList p; int i; L = (LinkList)malloc(sizeof(LNode); L-next = NULL; / 先建立一个带头结点的单链表 for (i=n; i0; -i) p = (LinkList)mallo

20、c(sizeof(LNode); / 生成新结点 coutp-data.expnum; coutp-data.coef; p-next = L-next; L-next = p; / 插入到表头 / cout 链表生成. next; cout ; while (L) cout ( data.coef ,; cout data.expnum next; cout next; while (q) s = q-next; q-next = p; p = q; q = s; L-next = p;/Reverse_L/-void Split_L(LinkList &plyn, LinkList &odd) LinkList p,q,s; odd = (LinkList)malloc(sizeof(LNode); odd-next= NULL; q=ply