POJ

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.

BFS，三个方向bfs搜索，控制下一步的大小不要越界..

#include <iostream>#include <sstream>#include <cstdlib>#include <cstdio>#include <vector>#include <queue>#include <deque>#include <stack>#include <list>#include <map>#include <set>#include <climits>#include <ctime>#include <complex>#include <cmath>#include <string>#include <cctype>#include <cstring>#include <algorithm>using namespace std;typedef pair<int,int> pii;typedef long long ll;#define maxn 100005int dx[]= {0,0,-1,1};int dy[]= {1,-1,0,0};int n,k;int s[maxn];int vis[maxn];int bfs() {    queue <int> q;    q.push(n);    vis[n]=true;    s[n]=0;    while(!q.empty()) {        int t=q.front();        q.pop();        for(int i=0; i<3; i++) {            int tt=0;            if(i==0)                tt=t-1;            else if(i==1)                tt=t+1;            else                tt=t*2;            if(tt>100000||tt<0)                continue;            if(!vis[tt]) {                s[tt]=s[t]+1;                q.push(tt);                vis[tt]=true;                if(tt==k)                    return s[tt];            }        }    }    return 0;}int main() {    while(cin>>n>>k) {        memset(vis,0,sizeof(vis));        if(n>=k) {            cout<<n-k<<"\n";        } else            cout<<bfs()<<"\n";    }    return 0;}