poj 3278 Catch That Cow

题目链接：http://poj.org/problem?id=3278

题目描述：

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.

题目大意：

给定一个数N (O≤N≤100000)，变成另一个数K(O≤K≤100000)，允许的操作是乘以2，或者加减1，问最少要几步才能完成？

AC代码：

#include<cstdio>

#include<iostream>

#include<cstring>

using namespace std;

int n,k;

int q[400005],visit[400005],count[400005];

int BFS(int m)

{

    int front=0,rear=1;

    q[front]=m;

    count[m]=0;

    visit[m]=1;

    while(front<rear)

    {

        m=q[front++];

        if(!visit[m*2]&&m<k)

        {

            q[rear++]=m*2;

            visit[m*2]=1;

            count[m*2]=count[m]+1;

            if(m*2==k)

                return count[m*2];

        }

        if(!visit[m+1]&&m<k)

        {

            q[rear++]=m+1;

            count[m+1]=count[m]+1;

            visit[m+1]=1;

            if(m+1==k)

                return count[m+1];

        }

        if(!visit[m-1]&&m>0)

        {

            q[rear++]=m-1;

            count[m-1]=count[m]+1;

            visit[m-1]=1;

            if(m-1==k)

                return count[m-1];

        }

    }

    return -1;

}

 int main()

{

    while(scanf("%d%d",&n,&k)!=EOF)

    {

        memset(visit,0,sizeof(visit));

        memset(count,0,sizeof(count));

        if(n>=k)

            printf("%d\n",n-k);

        else

        {

            printf("%d\n",BFS(n));

        }

    }

    return 0;

}