poj 3278 Catch That Cow

Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 71890 Accepted: 22628
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
大致题意：
给定两个整数n和k
通过 n+1或n-1 或n*2 这3种操作，使得n==k
输出最少的操作次数

/*
BFS状态扩展   
{
    只要找到解就一定是最优的.
    同样这题我们也可以用DFS跑回溯（显然要爆T）
    其实迭代搜也能跑（还不会orz.） 
} 
*/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#define MAXN 1000001
using namespace std;
int n,k,s[MAXN];
bool b[MAXN];
queue<int>qq;
int bfs()
{
    qq.push(n);
    s[n]=0;
    b[n]=true;
    if(n>=k) return n-k;
    while(!qq.empty())
    {
        int x=qq.front(),y;
        qq.pop();
        for(int i=1;i<=3;i++)
        {
            if(i==1) y=x-1;
            if(i==2) y=x+1;
            if(i==3) y=x*2;
            if(y<0||y>MAXN) continue;
            if(!b[y])
            {
                s[y]=s[x]+1;
                b[y]=true;
                qq.push(y);
            }
            if(y==k) return s[y];
        }
    }
}
int main()
{
    cin>>n>>k;
    printf("%d",bfs());
    return 0;
}