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
思路：
这个题目又开发广搜的一个新的功能：求出图中两点的最短距离
这个证明是通过搜索树得到的，想象一个点可以扩展为其他三个点，那么被扩展出来的这三个点到最初始点的那个距离就是最短的，因为如果不是最短的，那么这三个点在之前就会出现
其实质就是一个最优路线上的每一个被扩展出的点都是最优的。
#include <iostream>
#include <cstring>
#include <queue>
#define MAX 100007
#define INF 0x7fffffff
using namespace std;
int n,k;
int vis[MAX];
int dis[MAX];
int bfs()
{
queue<int> q;
q.push(n);
int next;
while(!q.empty())
{
int tmp = q.front();
q.pop();
for(int i = 1;i <= 3;i++) {
if(i == 1) next = tmp-1;
else if(i == 2) next = tmp+1;
else next = tmp*2;
if(next<0 || next>MAX)
continue;
if(!vis[next]) {
vis[next] = 1;
dis[next] = dis[tmp]+1;
q.push(next);
}
if(next == k)
return dis[k];
}
}
}
int main()
{
while(cin>>n>>k) {
memset(vis,0,sizeof(vis));
vis[n] = 1;
if(n >= k)
cout<<n-k<<endl;
else cout<<bfs()<<endl;
}
return 0;
}
原文：http://www.cnblogs.com/immortal-worm/p/5136452.html