hdu2717:Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14813    Accepted Submission(s): 4476

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
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 minute

思路：当前所在的点看作根节点，那么可以走的三种路径就是三个儿子。这样就很容易看出这是一个树，bfs

#include<iostream>#include<cstdio>#include<string>#include<map>#include<algorithm>#include<string.h>#include<queue>#define ll long longusing namespace std;int n,k;int v[200005];int bfs(){    int a;    a=n;    v[n]=1;    queue<int>q;    q.push(a);    while(!q.empty()){        a=q.front();        q.pop();        if(a==k){            return v[a];        }        if(!v[a-1]&&a-1>=0){            v[a-1]=v[a]+1;            q.push(a-1);        }        if(!v[a+1]&&a+1<=100000){            v[a+1]=v[a]+1;            q.push(a+1);        }        if(!v[a*2]&&a*2<=100000){//数组要开大点 v[200005]，不然会越界 or a*2放在 !v[a*2]前面            v[a*2]=v[a]+1;            q.push(a*2);        }    }    return -1;}int main(){    while(cin>>n>>k){        memset(v,0,sizeof(v));        int ans;        ans=bfs();        cout<<ans-1<<endl;    }}