uva 133 The Dole Queue

//算法描述，构建一个双向循环链表，不需要不存放数据的头指针，按照prior指针遍历则认为是逆时针，按照next指针遍历则认为是顺时针

 

//prior是逆时针，next是顺时针

#include <stdio.h>

#include <stdlib.h>

#define LEN sizeof(struct node)

int N,k,m;

struct node

{

    int num;

    struct node *prior,*next;

};

void print_link(struct node *L)

{

    struct node *p;

    p=L; printf("%d\n",p->num); p=p->prior;

    while(p!=L) {printf("%d\n",p->num);p=p->prior;}

    printf("***************************\n");

    p=L; printf("%d\n",p->num); p=p->next;

    while(p!=L) {printf("%d\n",p->num);p=p->next;}

    printf("***************************\n");

}



int main()

{

    struct node *L,*l1,*l2,*temp1,*temp2;  int i;  int flag,count;  int m1,m2;  

    while(scanf("%d%d%d",&N,&k,&m)!=EOF && N && k && m )

    {

        L=l1=(struct node*)malloc(LEN);  L->prior=L->next=NULL; L->num=1;

        for(i=2; i<=N; i++) 

        {

            l2=(struct node*)malloc(LEN); l2->num=i;

            l1->prior=l2; l2->next=l1; l2->prior=NULL; l1=l2;

        }

        l1->prior=L; L->next=l1; l1=L; l2=L->next;  count=0;  

//        printf("%d %d\n",l1->num,l2->num);  //print_link(L);

        flag=0;

        while(count<(N-2) )

        {

            for(i=1; i<k; i++)  l1=l1->prior;  m1=l1->num; 

            for(i=1; i<m; i++)  l2=l2->next;   m2=l2->num; 

            if(m1==m2)  

            {

                count++;  if(flag)  printf(",");  

                temp1=l1->prior;  temp2=l1->next;   temp1->next=temp2;  temp2->prior=temp1; 

                printf("%3d",l1->num); 

                free(l1);  l1=temp1; l2=temp2;

                flag=1;

            }

            else

            {

                count+=2;   if(flag) printf(",");  

                temp1=l1->prior;  temp2=l1->next;  temp1->next=temp2;  temp2->prior=temp1; 

                printf("%3d",l1->num); 

                free(l1);  l1=temp1;  if(l1==l2) l1=l1->prior;  //这个判断很重要否则会有BUG

                temp1=l2->prior;  temp2=l2->next;  temp1->next=temp2;  temp2->prior=temp1; 

                printf("%3d",l2->num); 

                free(l2);  l2=temp2;

                flag=1;

            }

        }



        if(count==(N-1))   //不能漏掉这种情况否则程序会奔溃掉

        {  if(flag) printf(",");  printf("%3d\n",l1->num);  continue;  }

        for(i=1; i<k; i++)  l1=l1->prior;  m1=l1->num; 

        for(i=1; i<m; i++)  l2=l2->next;   m2=l2->num;



        if(m1==m2)

        {

            if(flag) printf(",");  printf("%3d",l1->num); 

            l2=l1->prior; printf(","); printf("%3d",l2->num); printf("\n"); 

        }

        else 

        {

            if(flag) printf(",");  printf("%3d%3d\n",l1->num,l2->num); 

        }

        free(l1); free(l2);  

    }

    return 0;

}