原来这个叫做广搜啊。。。

题目来源  poj3278 ；

 Catch That Cow

Time Limit: 2000MSMemory Limit: 65536K

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 fastes  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.

这个题目的思想还是挺简单，也不要求剪枝优化什么的，所以根据题意直接写，另外，通过这个题目学了一些stl <queue>库函数的一些用法。。

代码：

#include <iostream>#include <queue>#include <cstdio>#include <cstring>using namespace std;int main(){int line[100001]={0};bool sign[100001]={0};queue<int>x;        // 创建一个int型的队列。。int n,k;cin>>n>>k;    x.push(n);       // 将n压入队列x队首。    line[n]=0;           while(!x.empty()){  int y=x.front()  ;     //将x的队头赋值给y。;  x.pop();              //上一步没有将队头取出，这一步取出队头。  sign[n]=true;         //初始位置设为走过。  if(y==k) break;          else   {if(y>=1&&!sign[y-1]) {x.push(y-1);             //将y-1压入x队首。line[y-1]=line[y]+1;sign[y-1]=true; }if(y+1<=100000&&!sign[y+1])  {x.push(y+1);line[y+1]=line[y]+1;sign[y+1]=true;  }if(y<=50000&&!sign[y*2])  {x.push(y*2);line[y*2]=line[y]+1;sign[y*2]=true;  }  }}printf("%d\n",line[k]);return 0;}