POJ 3278 Catch That Cow

Farmer 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?

/*****************************************************/
/****    POJ 3278 Catch That Cow                ******/
/****    Wangguoliang   @Greenday               ******/
/****    2014-5-12                              ******/

#include<stdio.h>
#include<iostream>
#include<queue>
#include<string.h>	
using namespace std;
int visit[200001];//1为遍历过，0为未遍历 
int n,m,flag;
struct node  //定义结点 
{	 
   int pi;
   int step;
};
queue<node>Q;//定义对列 Q
void BFS()        //广度优先搜索 
{
    node p,q;
    int i,j;
    p.pi=n,p.step=0;
    Q.push(p);    //入队 
    while(!Q.empty())//判断是否队列已空 
    {
        p=Q.front();  //返回队列第一个元素 
        Q.pop();      //函数删除队列的一个元素
        if(p.pi==m)  //找到终点，结束
        {
            flag=p.step;
            return ;
        }
        p.step++;      //未找到步加1
        if(p.pi<m&&!visit[p.pi]) //直接乘以2
        {
            q=p;
            q.pi*=2;
            Q.push(q);
        }
        if(p.pi>0&&!visit[p.pi])//判断是否大于0，然后--
        {
            q=p;
            q.pi--;
            Q.push(q);
        }
        if(!visit[p.pi])//直接++，然后把这个点标记为遍历过 
        {
            q=p;
            q.pi++;
            Q.push(q);
            visit[p.pi]=1;
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(visit,0,sizeof(visit));//去除标记清0 
        flag=0;
        BFS();
        printf("%d\n",flag);
        //while(!Q.empty())//清空队列
          Q.pop();
    }
    return 0;
}
/*
#include<iostream>
#include<queue>
using namespace std;
int a[100001],b[100001];
int main()
{
   int m,n,y;
   queue<int>x;//建立名为x的队列
   cin>>m>>n;
   x.push(m);//在队列后加入m
   a[m]=0;
  while(x.size()!=0)//队列大小不为0
  {
  y=x.front();//y取队列首位
  x.pop();//去掉队列首位
  b[y]=1;
  if(y==n)
   break;
  if(y-1>=0&&b[y-1]==0)
  {
   x.push(y-1);
   a[y-1]=a[y]+1;
   b[y-1]=1;
  }
  if(y+1<=100000&&b[y+1]==0)
  {
   x.push(y+1);
   a[y+1]=a[y]+1;
   b[y+1]=1;
  }
  if(2*y<=100000&&b[2*y]==0)
  {
   x.push(2*y);
   a[2*y]=a[y]+1;
   b[2*y]=1;
  }
 }
 cout<<a[y]<<endl;
 return 0;
}
/*
//搜索的百度的不用stl 的算法
memory   980K  time 16Ms
#include <iostream>   
#include <cstring>   
#include <cmath>   
#include <cstdlib>   
#include <algorithm>   
#include <string>   
#include <cstdio>   
#include <climits>   
using namespace std;   
bool data[100005] = {0};   
int que[100005] = {0};   
int tnum[100005] = {0};   
void bfs(int n, int k)   
{   
   int front = 0, rear = 0;   
    que[0] = n;   
    data[n] = true;   
    tnum[0] = 0;   
    rear++;   
       
    while (front != rear)   
    {   
        int t = que[front];   
        int tn = tnum[front];   
        if (t == k)   
        {   
            printf("%d\n", tn);   
            return;   
        }   
           
        if (t > 0 && !data[t-1])   
        {   
            data[t-1] = true;   
            que[rear] = t-1;   
            tnum[rear] = tn+1;   
            rear++;   
        }   
           
        if (t < 100000 && !data[t+1])   
        {   
            data[t+1] = true;   
            que[rear] = t+1;   
            tnum[rear] = tn+1;   
            rear++;   
        }   
        if (t <= 50000 && !data[t*2])   
        {   
            data[t*2] = true;   
            que[rear] = t*2;   
            tnum[rear] = tn+1;   
            rear++;   
        }   
        front++;   
    }   
    printf("0\n");   
}   
       
int main()   
{   
    int n, k;   
    scanf("%d%d", &n, &k);   
    memset(data, 0, sizeof(data));   
    bfs(n, k);   
    return 0;   
}