poj 3278 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?

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos(-1.0);
typedef long long ll;

const int INF=1000010;

using namespace std;

int a[N],b[N];

queue<int>mp;

int bfs(int k,int n)
{
    mp.push(k);
    b[k]=1;
    int mans;
    while(mp.size())
        {
            mans=mp.front();
            mp.pop();
            if(mans==n)
                break;
            if((mans-1)>=0&&!b[mans-1])
                {
                    mp.push(mans-1);
                    a[mans-1]=a[mans]+1;
                    b[mans-1]=1;
                }
            if((mans+1)<=100000&&!b[mans+1])
                {
                    mp.push(mans+1);
                    a[mans+1]=a[mans]+1;
                    b[mans+1]=1;
                }
            if((mans*2)<=100000&&!b[mans*2])
                {
                    mp.push(mans*2);
                    a[mans*2]=a[mans]+1;
                    b[mans*2]=1;
                }
        }
        return a[n];
}
int main()
{
    int k,n;
    while(cin>>k>>n)
        {
            if(n<=k)
                printf("%d\n",k-n);
            else
                {
                    memset(a,0,sizeof(a));
                    memset(b,0,sizeof(b));
                    printf("%d\n",bfs(k,n));
                }
        }
    return 0;
}