(poj)Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 78708 Accepted: 24827

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 <string.h>#include <queue>using namespace std;struct node{    int data,step;};int vis[200000];void func(int n,int k){    struct node t,v;    queue<struct node>S;    t.step=0;    t.data=n;    vis[n]=1;    S.push(t);    while(!S.empty())    {        v=S.front();        S.pop();        if(v.data==k)        {            cout<<v.step<<endl;            return ;        }        if(v.data>=1&&!vis[v.data-1])        {            t.data=v.data-1;            t.step=v.step+1;            S.push(t);            vis[v.data-1]=1;        }        if(v.data<=k&&!vis[v.data+1])        {            t.data=v.data+1;            t.step=v.step+1;            S.push(t);            vis[v.data+1]=1;        }        if(v.data<=k&&!vis[v.data*2])        {            t.data=v.data*2;            t.step=v.step+1;            S.push(t);            vis[v.data*2]=1;        }    }}int main(){    int n,k;    cin>>n>>k;    memset(vis,0,sizeof(vis));    func(n,k);    return 0;}