（广搜）Catch That Cow -- poj -- 3278

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.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>

using namespace std;

#define N 110000

struct node
{
    int x, step;
};

int s, e;
bool vis[N];

int BFS(int s)
{
    node p, q;
    p.x = s, p.step = 0;

    memset(vis, false, sizeof(vis));
    vis[s] = true;
    queue<node>Q;
    Q.push(p);

    while(Q.size())
    {
        p = Q.front(), Q.pop();

        if(p.x == e) return p.step;

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

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

    return -1;
}

int main()
{
    while(scanf("%d%d", &s, &e)!=EOF)
    {
        int ans = BFS(s);

        printf("%d\n", ans);
    }
    return  0;
}