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)+);
     strcpy(p->name, name);
     p->sex    = sex;
     p->father = NULL;
     p->mother = NULL;
     p->mate   = NULL;
     for(index=; 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=; index<CHILDREN-; 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=; index<CHILDREN-; 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 ;
     if(p1->sex==p2->sex) return ;/*不可能是堂兄妹*/
     f1 = p1->father;
     f2 = p2->father;
     if(f1!=NULL && f1==f2) return ;/*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 ;
     }
     return ;
 }

 /* 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("\n");
     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=; index<CHILDREN-; index++)
             if(p->childern[index]==NULL)
                 break;
         if(index>)
             printf(" Children are:");
         for(i=; i<index; i++)
             printf(" %s",p->childern[i]->name);
     }
     printf("\n");
 }

 int main()
 {
     char *name[] = {"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[],male);
     pFather1     = newperson(name[],male);
     pFather2     = newperson(name[],male);
     pMother1     = newperson(name[],female);
     pMother2     = newperson(name[],female);
     pSon         = newperson(name[],male);
     pDaughter    = newperson(name[],female);
     pCousin      = newperson(name[],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.\n",pDaughter->name,pCousin->name);
     else
         printf("%s and %s are brothers (sisters) in law.\n",pDaughter->name,pCousin->name);
     if(!brothersinlaw(pSon,pCousin))
         printf("%s and %s are not brothers (sisters) in law.\n",pSon->name,pCousin->name);
     else
         printf("%s and %s are brothers (sisters) in law.\n",pSon->name,pCousin->name);
     if(!brothersinlaw(pSon,pDaughter))
         printf("%s and %s are not brothers (sisters) in law.\n",pSon->name,pDaughter->name);
     else
         printf("%s and %s are brothers (sisters) in law.\n",pSon->name,pDaughter->name);

     //printf("Hello world!\n");
     return ;
 }