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?

Line 1: Two space-separated integers: N and K

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

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. 

#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
#include<queue>
#include<stack>
using namespace std;
#define N 100010
int time[N], maps[N], s, e;//将这些宏定义是为了避免传参；
void BFS();
int main()
{
while(scanf("%d%d", &s, &e)!=EOF)
{
memset(time, 0, sizeof(time));//定义的时间数组，初始化为零；本身定义了结构体，其中包含了位置location、时间time、标记变量flag。却发现flag无法初始化
memset(maps, 0, sizeof(maps));//标记是否走过
if(s==e)
printf("0\n");
else
BFS();
}
return 0;
}
void BFS()
{
queue<int>que;
int i, x, dir;
maps[s]=1;
que.push(s);
while(!que.empty())
{
x=que.front();
que.pop();
for(i=0; i<3; i++)
{
switch(i)//用switch更简便；
{
case 0: dir=x-1; break;
case 1: dir=x+1; break;
case 2: dir=x*2; break;
}
if(dir==e)//结束条件。其实也可放在for循环外，只不过是多进行了几步而已
{
printf("%d\n", time[x]+1);
return;
}
if(maps[dir]==0&&dir<N&&dir>=0)//这也算是剪枝，减少了重复
{
que.push(dir);
maps[dir]=1;
time[dir]=time[x]+1;
}
}
}
}