7-49 打印学生选课清单 (25分)

 

 

 

 

 

 解题思路：（此前用哈希表存储学生选课信息，最后一个测试点超时，或者内存超限）

后在网上翻看其他大能写的文章，受益颇多

注意到学生姓名的组成是3个大写字母+1个数字，故可开辟一个四维结构体数组指针，使学生姓名映射到唯一地址，也省去了用哈希函数要解决冲突的时间消耗问题

解法一、将学生的选课清单用二叉排序树结构存储，中序遍历输出即可

#include 
#include 

typedef struct TNode {
    int CourseNo;
    struct TNode *Left,*Right;
}*Tree;

typedef struct Node {
    int num;
    struct TNode *Next;
} Node;

typedef char Element[5];

Tree BuildBiSearchTree(Tree T,int CourseNo);
void Trav(Tree T);
void Out(Node *head);

int main(int argc,char **argv) {
    int i,j,k,t;
    int n,m;
    int CourseNo,Num;
    Element Name;

    Node *List[26][26][26][10];
    for(i=0; i<26; i++) {
        for(j=0; j<26; j++) {
            for(k=0; k<26; k++) {
                for(t=0; t<10; t++) {
                    List[i][j][k][t]=NULL;
                }
            }
        }
    }
    scanf("%d %d",&n,&m);
    for(i=0; i) {
        scanf("%d %d",&CourseNo,&Num);
        for(j=0; j) {
            scanf("%s",Name);
            if(List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0']==NULL)
            {
                List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0']=malloc(sizeof(struct Node));
                List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0']->num=0;
                List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0']->Next=NULL;
            }
            List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0']->num++;
            List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0']->Next
                =BuildBiSearchTree(List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0']->Next,CourseNo);
        }
    }

    for(i=0; i) {
        scanf("%s",Name);
        printf("%s",Name);
        if(List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0']!=NULL)
        Out(List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0']);
        else
        printf(" 0");
        printf("\n");
    }
}

Tree BuildBiSearchTree(Tree T,int CourseNo) {
    if(!T) {
        T=malloc(sizeof(struct TNode));
        T->CourseNo=CourseNo;
        T->Left=T->Right=NULL;
    } else if(CourseNoCourseNo)
        T->Left=BuildBiSearchTree(T->Left,CourseNo);
    else
        T->Right=BuildBiSearchTree(T->Right,CourseNo);
    return T;
}
void Trav(Tree T) {
    if(T) {
        Trav(T->Left);
        printf(" %d",T->CourseNo);
        Trav(T->Right);
    }
}
void Out(Node *head) {
    printf(" %d",head->num);
    if(head->Next) {
        Tree T=head->Next;
        Trav(T);
    }

}

 

 

解法二、将学生的选课清单用单链表结构存储（插入排序）

#include 
#include 

typedef struct Node {
    int Course;
    struct Node *Next;
}Node;

typedef char Element[5];
Node *Insert(Node *sort,int k);
void Out(Node *head,int N);

int main(int argc,char **argv) {
    int i,j,k,t;
    int n,m;
    int CourseNo,Num;
    Element Name;
    
    Node *List[26][26][26][10];
    for(i=0; i<26; i++) {
        for(j=0; j<26; j++) {
            for(k=0; k<26; k++) {
                for(t=0; t<10; t++) {
                    List[i][j][k][t]=NULL;
                }
            }
        }
    }
    scanf("%d %d",&n,&m);
    for(i=0;i)
    {
        scanf("%d %d",&CourseNo,&Num);
        for(j=0;j)
        {
            scanf("%s",Name);
            List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0']
            =Insert(List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0'],CourseNo);
        }
    }
    
    for(i=0;i)
    {
        scanf("%s",Name);
        printf("%s",Name);
        Out(List[Name[0]-'A'][Name[1]-'A'][Name[2]-'A'][Name[3]-'0'],0);
        printf("\n");
    }
}

Node *Insert(Node *sort,int k)
{
    if(!sort||sort->Course<k)
    {
        Node *n=malloc(sizeof(struct Node));
        n->Course=k;
        n->Next=sort;
        return n;
    }
    else if(sort->Course!=k)
    sort->Next=Insert(sort->Next,k);
    return sort;
}
void Out(Node *head,int N)
{
    if(head)
    {
        Out(head->Next,++N);
        printf(" %d",head->Course);
    }
    else
    printf(" %d",N);
}