C语言实例解析精粹学习笔记——36（模拟社会关系）

实例：

　　设计一个模拟社会关系的数据结构，每个人的信息用结构表示，包含名字、性别和指向父亲、母亲、配偶、子女的指针（只限两个子女）。要求编写以下函数：

　　　　（1）增加一个新人的函数

　　　　（2）建立人与人之间关系的函数：父-子、母-子、配偶等。

　　　　（3）检查两人之间是否为堂兄妹

思路解析：

　　能够充分的联系指针的应用。书中的代码在增加一个新人时，只为新人提供名字和性别，关于新人的其他信息通过调用其他函数建立。

书中代码如下：

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

#define CHILDREN 2

struct person{
    char *name;              /*名字符串指针*/
    char sex;                /*性别：男用字符'M'；女用字符'F'*/
    struct person *father;   /*指向父亲*/
    struct person *mother;   /*指向母亲*/
    struct person *mate;     /*指向配偶*/
    struct person *childern[CHILDREN];/*指向子女*/
};

/* [函数]newperson增加新人 */
struct person *newperson(char *name, char sex)
{
    struct person *p;
    int           index;

    p       = (struct person *)malloc(sizeof(struct person));
    p->name = (char *)malloc(strlen(name)+1);
    strcpy(p->name, name);
    p->sex    = sex;
    p->father = NULL;
    p->mother = NULL;
    p->mate   = NULL;
    for(index=0; index<CHILDREN; index++)
        p->childern[index] = NULL;
    return p;
}

/* [函数]father_child建立父－子关系 */
void father_child(struct person *father, struct person *child)
{
    int index;

    for(index=0; index<CHILDREN-1; index++)/*寻找一个空缺的子女指针*/
        if(father->childern[index]==NULL)  /*若没有空缺，则填在最后*/
            break;
    father->childern[index] = child;    /*建立父－子关系*/
    child->father = father;
}

void mother_child(struct person *mother, struct person *child)
{
    int index;
    for(index=0; index<CHILDREN-1; index++)/*寻找一个空缺的子女指针*/
        if(mother->childern[index]==NULL)  /*若没有空缺，则填在最后*/
            break;
    mother->childern[index] = child;    /*建立母－子关系*/
    child->mother = mother;
}

/* [函数]mate 建立配偶关系 */
void mate(struct person *h, struct person *w)
{
    h->mate = w;
    w->mate = h;
}

/* [函数]brotherinlow 检查两人是否是堂兄妹 */
int brothersinlaw(struct person *p1, struct person *p2)
{
    struct person *f1, *f2;
    if(p1==NULL||p2==NULL||p1==p2)
        return 0;
    if(p1->sex==p2->sex) return 0;/*不可能是堂兄妹*/
    f1 = p1->father;
    f2 = p2->father;
    if(f1!=NULL && f1==f2) return 0;/*是兄妹，不是堂兄妹*/
    while(f1!=NULL&&f2!=NULL&&f1!=f2)/*考虑远房情况*/
    {
        f1 = f1->father;
        f2 = f2->father;
        if(f1!=NULL && f2!=NULL && f1==f2) return 1;
    }
    return 0;
}

/* 函数print_relate用于输出人物p的姓名，性别和各种关系 */
void print_relate(struct person *p)
{
    int index,i;
    if(p->name == NULL)
        return;
    if(p->sex == 'M')
        printf(" %s is male.", p->name);
    else
        printf(" %s is female.",p->name);
    if(p->father != NULL)
        printf(" %s's father is %s.",p->name,p->father->name);
    if(p->mother != NULL)
        printf(" %s's mother is %s.",p->name,p->mother->name);
    printf("
");
    if(p->mate != NULL)
        if(p->sex == 'M')
            printf(" His wife is %s.", p->mate->name);
        else
            printf(" Her husband is %s.",p->mate->name);
    if(p->childern != NULL)
    {
        for(index=0; index<CHILDREN-1; index++)
            if(p->childern[index]==NULL)
                break;
        if(index>0)
            printf(" Children are:");
        for(i=0; i<index; i++)
            printf(" %s",p->childern[i]->name);
    }
    printf("
");
}

int main()
{
    char *name[8] = {"John","Kate","Maggie","Herry","Jason","Peter","Marry","Jenny"};
    char male='M', female='F';
    struct person *pGrandfather, *pFather1, *pFather2, *pMother1, *pMother2;
    struct person *pSon, *pDaughter, *pCousin;

    pGrandfather = newperson(name[0],male);
    pFather1     = newperson(name[3],male);
    pFather2     = newperson(name[4],male);
    pMother1     = newperson(name[1],female);
    pMother2     = newperson(name[2],female);
    pSon         = newperson(name[5],male);
    pDaughter    = newperson(name[6],female);
    pCousin      = newperson(name[7],female);

    father_child(pGrandfather,pFather1);
    father_child(pGrandfather,pFather2);
    father_child(pFather1,pSon);
    father_child(pFather1,pDaughter);
    father_child(pFather2,pCousin);

    mate(pFather1,pMother1);
    mate(pFather2,pMother2);
    mother_child(pMother1,pSon);
    mother_child(pMother1,pDaughter);
    mother_child(pMother2,pCousin);

    print_relate(pGrandfather);
    print_relate(pFather1);
    print_relate(pFather2);
    print_relate(pMother1);
    print_relate(pMother2);
    print_relate(pSon);
    print_relate(pDaughter);
    print_relate(pCousin);


    if(!brothersinlaw(pDaughter,pCousin))
        printf("%s and %s are not brothers (sisters) in law.
",pDaughter->name,pCousin->name);
    else
        printf("%s and %s are brothers (sisters) in law.
",pDaughter->name,pCousin->name);
    if(!brothersinlaw(pSon,pCousin))
        printf("%s and %s are not brothers (sisters) in law.
",pSon->name,pCousin->name);
    else
        printf("%s and %s are brothers (sisters) in law.
",pSon->name,pCousin->name);
    if(!brothersinlaw(pSon,pDaughter))
        printf("%s and %s are not brothers (sisters) in law.
",pSon->name,pDaughter->name);
    else
        printf("%s and %s are brothers (sisters) in law.
",pSon->name,pDaughter->name);

    //printf("Hello world!
");
    return 0;
}