poj3278-Catch That Cow

Problem Description
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.
Example Input

5 17

Example 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.

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#include <iostream>#include <cstdio>#include <algorithm>#include <queue>#include <cstring>using namespace std;int book[1000005];int n,k;bool judge(int x){    if(x<0||x>=1000000||book[x])        return 0;    return 1;}struct node{    int step;    int point;}t,v;void bfs(int n){    queue<node>q;    t.point=n;    t.step=0;    q.push(t);    book[n]=1;    while(!q.empty())    {        v=q.front();        q.pop();        if(v.point==k)        {            cout<<v.step<<endl;            return ;        }        t=v;        t.point=v.point+1;        if(judge(t.point))        {            t.step=v.step+1;            book[t.point]=1;            q.push(t);        }        t.point=v.point-1;        if(judge(t.point))        {            t.step=v.step+1;            book[t.point]=1;            q.push(t);        }        t.point=v.point*2;        if(judge(t.point))        {            t.step=v.step+1;            book[t.point]=1;            q.push(t);        }    }}int main(int argc, char const *argv[]){    scanf("%d%d",&n,&k);    bfs(n);    return 0;}