我的第一个小算法~（三叉树、队列）

题目
Problem DescriptionFarmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?  InputLine 1: Two space-separated integers: N and K OutputLine 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow. Sample Input5 17Sample Output4HintThe fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes
代码
#include"stdio.h"
#include"stdlib.h"
#include"math.h" 
typedef struct QNode
{
    int data; 
    struct QNode *next;
} QNode,*QueuePtr;   //QNode 队列节点这儿还用了typedef的符合语法，QueuePtr也被定义为类型别名，代表队列节点指针类型；
 
typedef struct 
{ 
    QNode*front; 
    QNode*rear;
}  LinkQueue;     //定义一个“句柄”结构体实例，指向队列节点的头与尾
 
voidInitQueue(LinkQueue &Q)//初始化队列，使句柄结构体实例指向一个队列
{ 
    Q.front=(QNode*)malloc(sizeof(QNode));
    if(!Q.front) exit(-1); //-1就是EXIT_FAILURE预编译器变量
    Q.front->next=NULL;
    Q.rear=Q.front;
}
voidEnQueue(LinkQueue &Q,int e)
{
    Q.rear->next=(QNode*)malloc(sizeof(QNode));
    Q.rear=Q.rear->next;
    Q.rear->data=e;
    Q.rear->next=NULL;
} 
voidDeQueue(LinkQueue &Q,int &e)
{ 
    if(!Q.front->next) 
       ;
    else 
    {
       QNode*p=Q.front->next; 
       e=p->data;
       Q.front->next=p->next;
       if(p==Q.rear) 
           Q.rear=Q.front;
       free(p);
    }
}
boolIsEmpty(LinkQueue Q)
{ 
    if(Q.front==Q.rear) 
       return true; return false;
} 
intMinLength(int start,intdes)//求start到des的最短路径的长度
{ 
    int n=0,e; 
    LinkQueueQ; 
    InitQueue(Q);
    EnQueue(Q,start);
    while(!IsEmpty(Q))//其实Q永远不会为空，这个循环每次让Q指向队列的节点加
    {  
       DeQueue(Q,e);
       n++;             //每取一个元素都加，n代表循环次数，也代表当前检查到三叉
       if(e!=des)       //数的节点位置，当检查到n的时候队列中共有n+1个节点，
       {                  //虚拟的三叉树此时有n+1个节点
           EnQueue(Q,e-1); 
           EnQueue(Q,e+1);
           EnQueue(Q,2*e);  
       }
/*
 
 //这后面的if else 语句是为了判定n是否恰好到了三叉树的某一层的最右边的
 //那个节点，如果是这样，那么n+1恰好是的x次幂，因为double存储有误差
 //所以如果一个double值取整跟它本身的值差e-6(或-（e-6）)就当它是整数  
 //此时算出的三叉树高度比n所在的高度刚好大一，其他情况算出的高度取整就是        
 //实际值，这个地方应该先加.1再取整,保证double取到最        
 //接近的整值
*/
       else if
           (fabs((int)(log(2*n+1)/log(3))-log(2*n+1)/log(3))<1e-6|| fabs( (int)(log(2*n+1)/log(3))-log(2*n+1)/log(3))>1-1e-6)   
           return (int)(log(2*n+1)/log(3)+0.1)-1;  
       else   
           return (int)(log(2*n+1)/log(3));//
    }
} 
voidmain()
{ 
    int start,des;
    scanf("%d%d",&start,&des);//输入起点和终点
    printf("%d",MinLength(start,des));
    ::system("pause");
}