POJ 3278 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.

题解：

　　BFS，注意剪枝。

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=1e5+5;
bool vis[maxn*2];
struct node
{
    int x,step;
};
int k;
int bfs(int n)
{
    queue<node> que;
    node now;
    now.x=n;
    now.step=0;
    que.push(now);
    while(que.size())
    {
        now=que.front();
        que.pop();
        node next;
        if(now.x==k)
            return now.step;
        if(now.x<k&&!vis[now.x+1])
        {
            next.x=now.x+1;
            vis[next.x]=true;
            next.step=now.step+1;
            que.push(next);
        }
        if(now.x>0&&!vis[now.x-1])
        {
            next.x=now.x-1;
            vis[next.x]=true;
            next.step=now.step+1;
            que.push(next);
        }
        if(now.x<k&&!vis[now.x*2])
        {
            next.x=now.x*2;
            vis[next.x]=true;
            next.step=now.step+1;
            que.push(next);
        }
    }
    return -1;
}
int main()
{
    int n;
    while(cin>>n>>k)
    {
        memset(vis,false,sizeof(vis));
        cout<<bfs(n)<<endl;
    }
    return 0;
}