POJ 3278（bfs）

Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 32679 Accepted: 10060
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.
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<queue>using namespace std;const int maxn=100001;int step[100000];int vis[100000];queue<int> q;int n,k;int bfs(int n,int k){    q.push(n);    step[n]=0;    vis[n]=1;    while(!q.empty()){        int head,next;        head=q.front();        q.pop();        for(int i=0;i<3;i++){            if(i==0)    next=head-1;            else if(i==1)    next=head+1;            else     next=head*2;            if(next<0 || next>maxn) continue;            if(!vis[next]){                step[next]=step[head]+1;                q.push(next);                vis[next]=1;                if(next==k) return step[next];            }         }    }}int main(){    while(cin >> n >> k){        while(!q.empty()) q.pop();        if(n>k) cout << n-k << endl;        else        cout <<bfs(n,k) << endl;    }    return 0;}