POJ3278廣搜

題目：

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.

//題意：從n到k,每一步有+1，-1，*2 這三種運算方式，求解從n到k的所需要的最小的步數
//方法：廣搜
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAX=100010;
int step[MAX],vis[MAX];
int n,k;
int bfs()
{
    int t,next;
    queue<int>q;
    q.push(n);
    step[n]=0;
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        for(int i=1;i<=3;i++)
        {
            if(i==1)next=t+1;
            if(i==2)next=t-1;
            else if(i==3)next=t*2;
            if(next<0||next>MAX)continue;//此處注意
            if(!vis[next])
            {
                vis[next]=1;
                q.push(next);
                step[next]=step[t]+1;
            }
            if(next==k)return step[k];
        }
    }
}
int main()
{
    memset(vis,0,sizeof(vis));
    cin>>n>>k;
    if(n>=k)cout<<n-k<<endl;
    else cout<<bfs()<<endl;
    return 0;
}