B - Catch That Cow（BFS）

简单、BFS，注意的是因为有-1，所以状态会有返回并反复将相同的状态装入队列的情况，为了增加效率，防止内存超出，应该标记每次所经历过的状态，防止对同一个状态的多次访问。

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<cstdio>#include<iostream>#include<algorithm>#include<queue>#include<cstring>using namespace std;int n, k;//int cnt = 0;int d[100005];int vis[100005];int bfs(void){    memset(d, -1, sizeof(d));    memset(vis, 0, sizeof(vis));    if(n == k)        return 0;    queue<int> que;    que.push(n);    d[n] = 0;    vis[n] = 1;    while(que.size())    {        int q = que.front();        que.pop();        if(q == k)            break;        if(2 * q < 100005 && vis[2*q] == 0)        {            d[2 * q] = d[q] + 1;            vis[2*q] = 1;            que.push(2*q);        }        if(q + 1 < 100005 && vis[q+1] == 0)        {            d[q+1] = d[q] +1;            vis[q+1] = 1;            que.push(q+1);        }        if(q - 1 >= 0 && vis[q-1] == 0)        {            d[q-1] = d[q] +1;            vis[q-1] =1;            que.push(q-1);        }    }    return d[k];}int main(){    cin>>n>>k;    int cnt = bfs();    cout<<cnt<<endl;    return 0;}