POJ 3278 BFS

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 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.

 

tips：广搜入门题目。感觉广搜比深搜容易理解一点，因为不用递归，递归太过于抽象，也难debug。

　　数轴上，要从n到k去，每次三种变换坐标的方法（n+1，n-1,n*2)

　　用队列储存n，然后出队n，进队三个新坐标，如此循环，每次进队都要判断新坐标和k是否相等，不相等就step[next] = step[head] + 1；相等则return step[next]。

 

#include <bits/stdc++.h>
using namespace std;
#define maxn 100001
queue<int>q;
int step[maxn];
int vis[maxn];

int bfs(int n, int k){
	int head,next;
	q.push(n);//进队
	step[n] = 0;//步数初值为0
	vis[n] = 1;//标记这个点已经走过了
	while(!q.empty()){
		head = q.front();//把队首（n)的值赋给head，这一步必须放在循环外面。不然三个坐标会尝试三次。
		q.pop();//出队 
		for(int i = 0; i < 3; i++){
			if(i == 0) next = head-1;
			else if(i == 1) next = head+1;
			else  next = 2*head;//三个新坐标
			if(next > maxn || next < 0) continue;//考虑越界
			if(!vis[next]){
				q.push(next);//next进队
				step[next] = step[head] + 1;//步数+1，ps，这里用的step[head]+1，在每次循环的时候，他的初始值是一样的。
				vis[next] = 1;//标记为旧点
			} 
		if(k == next) return step[next];
	}	
}
}

int main(){
	int n,k;
	cin>>n>>k;
	memset(vis,0,sizeof(vis));
	memset(step,0,sizeof(step));
	while(!q.empty()) q.pop();
	if(n>k) cout<<n-k<<endl;
	else cout<<bfs(n,k)<<endl;
	return 0;
}