***参考Catch That Cow（BFS）

Catch That Cow

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 67   Accepted Submission(s) : 22

Problem 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

 

 

Source

PKU

 

 

题意：牛逃跑了，人要去抓回来这只牛，他们在一条直线上 ，现在牛的坐标是K，人的坐标是N，牛呆在原位置不动。人有两种行进方式，步行和传送，步行可以向前或向后走一步，传送为到达人当前坐标*2 的位置。每行进一次用一分钟，问人最少需要几分钟可以抓到牛。

 

思路：

 

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

const int maxn=100000;

int vis[maxn+10];
int n,k;

struct node
{
    int x,c;
};

int BFS()
{
    queue<node> q;
    while(!q.empty())
        q.pop();
    memset(vis,0,sizeof(vis));
    node cur,next;
    cur.x=n,cur.c=0;
    vis[cur.x]=1;
    q.push(cur);
    while(!q.empty())
    {
        cur=q.front();
        q.pop();
        for(int i=0; i<3; i++)
        {
            if(i==0)
                next.x=cur.x-1;
            else if(i==1)
                next.x=cur.x+1;
            else
                next.x=cur.x*2;
            next.c=cur.c+1;
            if(next.x==k)
                return next.c;
            if(next.x>=0 && next.x<=maxn && !vis[next.x])
            {
                vis[next.x]=1;
                q.push(next);
            }
        }
    }
    return 0;
}

int main()
{

    freopen("1.txt","r",stdin);

    while(~scanf("%d%d",&n,&k))
    {
        if(n>=k)
        {
            printf("%d\n",n-k);
            continue;
        }
        printf("%d\n",BFS());
    }
    return 0;
}