1. 1. 1.     move a onto b
      在將a搬到b上之前，先將a和b上的積木放回原來的位置（例如：1就放回1的最開始位罝）
      2. move a over b
      在將a搬到b所在的那堆積木之上之前，先將a上的積木放回原來的位罝（b所在的那堆積木不動）
      3. pile a onto b
      將a本身和其上的積木一起放到b上，在搬之前b上方的積木放回原位
      4. pile a over b
      將a本身和其上的積木一起搬到到b所在的那堆積木之上

 

//用链表实现，没有任何算法可言，完全就是模拟，感觉用数组来做代码能更短

//由于用链表来做的原因，涉及到指针所以很多细节问题，实际上调试很久，指针的东西太久没写都忘记

//了，以后要适当复习

//时间0.012

 

 

#include 
     
      #include 
      
       #include 
       
        #define LEN1 sizeof(struct list)#define LEN2 sizeof(struct node)#define MAX 30  //最多有25个积木struct list{    int num;    struct list *prior,*next;    struct node *head;}*L;struct node{    int num;    struct node *next;};int n,a,b,ans[30];  //ans是在输出答案时保存答案的void print_link(){    struct list *p=L;  struct node *q;    while(p)    {        printf("%d\n",p->num); q=p->head;        while(q) {printf("%3d",q->num); q=q->next;} printf("\n");        p=p->next;    }}void create_link(){    struct list *l1,*l2;  int i;    L=l1=(struct list*)malloc(LEN1); L->prior=L->next=NULL; L->num=0;    L->head=(struct node*)malloc(LEN2);  L->head->next=NULL;  L->head->num=0;    for(i=1; i
        
         prior=l1; l1->next=l2; l2->next=NULL; l2->num=i; l1=l2;        l2->head=(struct node*)malloc(LEN2); l2->head->next=NULL; l2->head->num=i;    }}void move(struct list *s){    struct node *p=s->head;  struct list *l;    while(p && p->num!=a && p->num!=b)    {        p=s->head; l=L;         while(l->num!=p->num)             l=l->next;        s->head=p->next;         p->next=l->head;         l->head=p;         p=s->head;    }}//在收到一条指令后要先判断这条指令是否合法，非法的指令是指a和b在同一堆积木中void command_1()  //move a onto b 在將a搬到b上之前，先將a和b上的積木放回原來的位置（例如：1就放回1的最開始位罝）{    struct list *l1=L,*l2; struct node *h1;  int count;    while(l1)    {        h1=l1->head;  count=0;        while(h1)          { if(h1->num==a || h1->num==b)  count++; h1=h1->next; }        if(count==1)  { move(l1); /*printf("move is ok ");*/ }        else if(count==2) return ;        l1=l1->next;    }//    printf("\nmove is end\n");    l1=L;     while(l1)     {        if((l1->head) && (l1->head->num)==a)  break;        l1=l1->next;    }    l2=L;     while(l2)     {        if((l2->head) && (l2->head->num)==b)  break;        l2=l2->next;    }    h1=l1->head;     l1->head=h1->next;    h1->next=l2->head;     l2->head=h1;//    printf("command_1 is end\n");}void command_2() //move a over b 在將a搬到b所在的那堆積木之上之前，先將a上的積木放回原來的位罝（b所在的那堆積木不動）{    struct list *l1=L,*l2; struct node *h1;  int flag,count;    while(l1)    {        h1=l1->head;  count=flag=0;        while(h1)          {             if(h1->num==a)  {flag=1;count++;}            if(h1->num==b)  count++;            h1=h1->next;         }        if(count==1 && flag)  { move(l1); /*printf("move is ok ");*/ break;}        else if(count==2) return ;        l1=l1->next;    }//    printf("\nmove is end\n");//    l1=L; while(l1->head->num!=a) l1=l1->next;    l2=L;    while(l2)    {        flag=0; h1=l2->head; while(h1)  {
   if(h1->num==b) {flag=1; break;} h1=h1->next;}        if(flag) break;        l2=l2->next;    }    h1=l1->head;     l1->head=h1->next;    h1->next=l2->head;     l2->head=h1;//    printf("command_2 is end\n");}void command_3()  //pile a onto b 將a本身和其上的積木一起放到b上，在搬之前b上方的積木放回原位{    struct list *l1,*l2; struct node *h1,*h2;  int flag,count;    l1=L;    while(l1)    {        h1=l1->head;  count=flag=0;        while(h1)          {             if(h1->num==b)  {flag=1;count++;}            if(h1->num==a)  count++;            h1=h1->next;         }        if(count==1 && flag)  { move(l1); /*printf("move is ok ");*/ break;}        else if(count==2) return ;        l1=l1->next;    }//    printf("\nmove is over\n");//    l1=L; while(l1->head->num!=a) l1=l1->next;    l2=L;    while(l2)    {        flag=0; h1=l2->head;         while(h1)  {    if(h1->num==a) {flag=1; break;}   h1=h1->next;   }        if(flag) break;        l2=l2->next;    }    h2=h1->next;     h1->next=l1->head;     l1->head=l2->head;     l2->head=h2;//    printf("command_3 is end\n");}void command_4() //pile a over b 將a本身和其上的積木一起搬到到b所在的那堆積木之上{    struct list *l1,*l2; struct node *h1,*h2,*temp1,*temp2;  int flag;    l1=L;    while(l1)    {        flag=0; h1=l1->head; while(h1) {
   if(h1->num==a) {flag=1; break;} h1=h1->next;}        if(flag) break;        l1=l1->next;    }    l2=L;    while(l2)    {        flag=0; h2=l2->head; while(h2) {
   if(h2->num==b) {flag=1; break;} h2=h2->next;}        if(flag) break;        l2=l2->next;    }    if(l1==l2)  return ;    temp1=h1->next;     h1->next=l2->head;      l2->head=l1->head;      l1->head=temp1;//    printf("command_4 is end\n");}void print_answer(){    struct list *p=L;  struct node *q;  int i;    while(p)    {        printf("%d:",p->num); q=p->head; i=-1;        while(q) { ans[++i]=q->num; q=q->next; }        while(i>=0) printf(" %d",ans[i--]);  printf("\n");        p=p->next;    }}int main(){    int i;  char s1[20],s2[20];    scanf("%d",&n); create_link();  //print_link();    while(1)    {        scanf("%s",s1);  if(!strcmp(s1,"quit"))  break;        scanf("%d%s%d",&a,s2,&b);  //printf("%s %d %s %d\n",s1,a,s2,b);        if(a==b) continue;        if     ( !strcmp(s1,"move") && !strcmp(s2,"onto") )          { /*printf("command_1\n");*/command_1(); }        else if( !strcmp(s1,"move") && !strcmp(s2,"over") )          { /*printf("command_2\n");*/command_2(); }        else if( !strcmp(s1,"pile") && !strcmp(s2,"onto") )          { /*printf("command_3\n");*/command_3(); }        else if( !strcmp(s1,"pile") && !strcmp(s2,"over") )          { /*printf("command_4\n");*/command_4(); }//        print_answer();    }    print_answer();    return 0;}