poj 3278 Catch That Cow

                                               Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 60418 Accepted: 18842

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX+ 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 andK

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

#include <iostream>#include <cstring>#include <queue>#include <cstdio>using namespace std;struct node{    int step;    int num;} b,c;int n,k;bool vis[100001*2];int BFS(){    memset(vis,false,sizeof(vis));    b.step = n;    b.num = 0;    vis[n] = true;    queue<node >q;    q.push(b);    while(!q.empty())    {        b = q.front();        q.pop();        if(b.step == k)            return b.num;        if(b.step + 1 <= k && !vis[b.step+1])        {            c.step = b.step+1;            c.num = b.num+1;            vis[c.step] = true;            q.push(c);        }        if(b.step - 1 >= 0 && !vis[b.step-1])        {            c.step = b.step-1;            c.num = b.num+1;            vis[c.step] = true;            q.push(c);        }        if(b.step <<1  <= k<<1 && !vis[b.step<<1])        {            c.step = b.step<<1;            c.num = b.num+1;            vis[c.step] = true;            q.push(c);        }    }}int main(){    while(cin>>n>>k)        cout<<BFS()<<endl;    return 0;}