题目：
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 101653 | Accepted: 31771 |
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
Output
Sample Input
5 17
Sample Output
4
大致题意：一个人站在n位置，每次可以往前走1个单位，或者往后走一个单位，或者直接闪现到当前位置*2的地方。问最少需要多少步到达k位置。
思路：裸的BFS。
坑点：（至今想不明白为啥）把输出放到main就wa了，在bfs函数里面输出就A了。
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<stack>typedef long long ll;using namespace std;const int MAXN = 100000;int n,k; int vis[100010];struct node{int pos;int step;
};node now,next;int bfs(){queue<node> Q;now.pos = n , now.step = 0;vis[n] = 1;Q.push(now);while(!Q.empty()){now = Q.front();Q.pop();if(now.pos == k){cout<<now.step<<endl;return;} if(now.pos-1 >= 0 && !vis[now.pos-1]){next.pos = now.pos - 1;next.step = now.step + 1;vis[next.pos] = 1;Q.push(next);}if(now.pos+1 <= MAXN && !vis[now.pos+1]){next.pos = now.pos + 1;next.step = now.step + 1;vis[next.pos] = 1;Q.push(next);}if(2*now.pos <= MAXN && !vis[now.pos*2]){next.pos = now.pos*2;next.step = now.step + 1;vis[next.pos] = 1;Q.push(next);}}
}int main(){ios::sync_with_stdio(false);cin>>n>>k;memset(vis,0,sizeof(vis));bfs();return 0;
}