Catch That Cow （BFS）

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.

经典BFS

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<cstdlib>
using namespace std;
int n,k;
int vis[150002]={0};
int step[150002]={0};
queue<int> q;
int bfs(int n){
    int fir=n,tmp;
    q.push(fir);
    vis[fir]=1;
    step[fir]=0;
    while(!q.empty()){
        fir=q.front();
        q.pop();
        for(int i=0;i<3;i++){
            if(i==0) tmp=fir+1;
            if(i==1) tmp=fir-1;
            if(i==2) tmp=fir*2;
            if(tmp<0||tmp>=150000)continue;
            else if(!vis[tmp]){
                vis[tmp]=1;
                step[tmp]=step[fir]+1;
                if(tmp==k)return step[tmp];
                q.push(tmp);
            }
        }
    }
}
int main()
{
    while(!q.empty())
        q.pop();
    scanf("%d%d",&n,&k);
    if(n>=k)
        printf("%d\n",n-k);
    else
        printf("%d\n",bfs(n));
    return 0;
}