poj 3278 搜索

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

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.

 

题意：

       农民和奶牛各在一个地方，农民可以通过+1，-1，*2来移动，每走一步用时一分钟，问移动到奶牛所在的位置需要多久。

题解：

         广搜，分别进行+1，-1，*2操作，存入队列。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>

using namespace std;
queue<int>q;
int a,b;
bool vis[100002];
int step[100002];

int bfs()
{
    int u,v;
    q.push(a);
    step[a]=0;
    vis[a]=true;
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        for(int i=0;i<3;i++)
        {
            if(i==0) v=u+1;
            else if(i==1) v=u-1;
            else v=u*2;
            if(v>=100001||v<0) continue;
            if(!vis[v])
            {
                step[v]=step[u]+1;
                vis[v]=true;
                q.push(v);
            }
            if(v==b) return step[v];
        }
    }
    return -1;
}

int main()
{
    while(cin>>a>>b)
    {
        memset(step,0,sizeof(step));
        memset(vis,false,sizeof(vis));
        while(!q.empty()) q.pop();
        if(a>=b) printf("%d\n",a-b);
        else
        {
            int ans=bfs();
            cout<<ans<<endl;
        }
    }
    return 0;
}