POJ3278-Catch That Cow

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

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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;
}