Catch That Cow （BFS luo搜 + 剪枝）

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.

 

 

结构体 + 数组标记，M E 。（心态爆炸 27次）2717 1/26 （魔改下去，一定有办法A掉

 

 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#define mem(s,t) memset(s,t,sizeof(s))
#define ok return 0;
#define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
int vis[100000+5];
int main()
{
    TLE;
    int n,m,now,nx;
    while(cin>>n>>m)
    {
        if(n>=m)
            cout<<n-m<<endl;
        else
        {
            mem(vis,0);            vis[n]=1;
            queue<int>q;
            while(!q.empty())  q.pop();
            q.push(n);
            while(!q.empty())
            {
                now = q.front();
                q.pop();
                if(now==m){    cout<<vis[now]-1<<endl;   break;}
                nx = now + 1;
                if(nx<=100000 && !vis[nx])  q.push(nx),vis[nx]=vis[now]+1;

                nx = now - 1;
                if(nx<=100000 && !vis[nx] && nx>=0)  q.push(nx),vis[nx]=vis[now]+1;

                nx = now*2;
                if(nx<=100000 && !vis[nx])  q.push(nx),vis[nx]=vis[now]+1;
            }
        }
    }
    ok;
}