poj---3278---Catch That Cow

题目链接： http://poj.org/problem?id=3278

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 minutes.
Source

USACO 2007 Open Silver

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include<vector>
#include<queue>
#include<algorithm>

using namespace std;
typedef long long LL;

const int maxn=500009;
const int INF=0x3f3f3f3f;
const int mod=2009;

struct node
{
    int x, step;
};
int n, k;
int vis[maxn];

int bfs()
{
    memset(vis, 0, sizeof(vis));
    queue<node>Q;
    node p, q;
    p.x=n;
    p.step=0;
    vis[p.x]=1;
    Q.push(p);

    while(Q.size())
    {
        q=Q.front();
        Q.pop();

        if(q.x==k)return q.step;

        for(int i=0; i<3; i++)
        {
            if(i==0)
                p.x=q.x+1;
            else if(i==2)
                p.x=q.x-1;
            else
                p.x=q.x*2;

            if(p.x>=0&&p.x<=100000&&!vis[p.x])
            {
                vis[p.x]=1;
                p.step=q.step+1;
                Q.push(p);
            }
        }
    }
    return -1;
}

int main()
{
    while(~scanf("%d %d", &n, &k))
    {
        int ans=bfs();
        printf("%d\n", ans);
    }
    return 0;
}