HDU 2717 Catch That Cow（bfs）

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?
Input
Line 1: Two space-separated integers: N and K
Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
Sample Input
5 17
Sample Output
4
Hint
The 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.

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;

typedef struct{
  int p;
  int step;
}node;
int book[100005];
queue<node> q;
int n, k;

int main()
{
  scanf("%d%d", &n, &k);
  if (n==k){
    printf("0");
    return 0;
  }
  while (!q.empty())
    q.pop();
  node t;
  t.p=n;
  t.step=0;
  book[n]=1;
  q.push(t);
  
  while (!q.empty()){
    node tmp=q.front();
    q.pop();
    
    int tx=tmp.p-1;
    if (tx>=0 && tx<=100000 && book[tx]==0){
      book[tx]=1;
      t.p=tx;
      t.step=tmp.step+1;
      q.push(t);
      if (t.p==k){
        printf("%d", t.step);
        return 0;
      }
    }
    
    tx=tmp.p+1;
    if (tx>=0 && tx<=100000 && book[tx]==0){
      book[tx]=1;
      t.p=tx;
      t.step=tmp.step+1;
      q.push(t);
      if (t.p==k){
        printf("%d", t.step);
        return 0;
      }
    }
    
    tx=tmp.p*2;
    if (tx>=0 && tx<=100000 && book[tx]==0){
      book[tx]=1;
      t.p=tx;
      t.step=tmp.step+1;
      q.push(t);
      if (t.p==k){
        printf("%d", t.step);
        return 0;
      }
    }
  }
  return 0;
}