POJ

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

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.

一维的bfs，每次有三种走法，x+1，x-1，x*2，bfs跑一遍输出步数就行.

#include <iostream>#include <cstdio>#include <queue>#include <algorithm>#include <cstring>using namespace std;typedef struct zb{    int x, step;}node;int v[999999];void bfs(int x, int k){    node a, b;    queue<node>q;    a.x = x;    a.step = 0;    v[a.x] = 1;    q.push(a);    while(!q.empty())    {        a = q.front();        q.pop();        if(a.x == k)        {            cout<<a.step<<endl;            break;        }        if(a.x + 1 <= k && !v[a.x+1])        {            b.x = a.x+1;            b.step = a.step + 1;            v[b.x] = 1;            q.push(b);        }        if(a.x * 2 <= 200000 && !v[a.x * 2])        {            b.x = a.x * 2;            b.step = a.step + 1;            v[b.x] = 1;            q.push(b);        }        if(a.x - 1 >= 0 && !v[a.x - 1])        {            b.x = a.x-1;            b.step = a.step + 1;            v[b.x] = 1;            q.push(b);        }    }}int main(){    int n, k;    cin>>n>>k;    memset(v,0,sizeof(v));    bfs(n,k);    return 0;}