问题描述：

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

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.

解题思路：

bfs，三个方向搜索，x=x+1,x=x-1,x=x*2，最先搜索到的就是用时最短的。

代码：

#include<cstdio>

#include<queue>

#include<cstring>

#include<iostream>

using namespace std;

int n,k;

int a[];

struct node

{

    int x;

    int step;

}now,net;

int bfs(int x)

{

    queue<node> q;

    now.x=x;

    now.step=;

    q.push(now);

    while(q.size())

    {

        now=q.front();

        q.pop();

        //三种情况

        if(now.x==k)return now.step;

        net.x=now.x+;

        if(net.x>=&&net.x<=&&a[net.x]==)

        {

            a[net.x]=;

            net.step=now.step+;

            q.push(net);

        }

        net.x=now.x-;

        if(net.x>=&&net.x<=&&a[net.x]==)

        {

            a[net.x]=;

            net.step=now.step+;

            q.push(net);

        }

        net.x=now.x*;

        if(net.x>=&&net.x<=&&a[net.x]==)

        {

            a[net.x]=;

            net.step=now.step+;

            q.push(net);

        }

    }

    return -;

}

int main()

{

    while(~scanf("%d%d",&n,&k))

    {

        memset(a,,sizeof(a));

        a[n]=;

        int ans=bfs(n);

        printf("%d\n",ans);

    }

    return ;

}

巴特西

Catch That Cow (BFS广搜)

问题描述：

解题思路：

代码：

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