Catch That Cow：BFS

Catch That Cow

Problem 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 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.

题目描述：找最短步数，用bfs

#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
int n, k,ans,step[200010],book[200010];		//wa了好几次，这里一定要大一点 ，step：步数 book：标记走没走 
void bfs(int a, int b) {
	ans = 0;
	book[a] = 1;
	queue<int >q;
	q.push(a);
	while (!q.empty()) {
		int x = q.front();
		//cout <<x<<"   "<<b<<"\n";
		q.pop();
		if( x== b)break;	
		
		if (x * 2 <= 200010 && x * 2 >= 0 && book[x * 2] == 0) {			//判断边界，判断走没走， 
			book[2 * x] = 1;
			q.push(x * 2);
			step[x * 2] = step[x] + 1;
		}
		if (x + 1 <= 200010 && x + 1 >= 0 && book[x + 1] == 0) {			//判断边界，判断走没走，
			book[1 + x] = 1;
			q.push(x +1);
			step[x + 1] = step[x] + 1;
		}			
		if (x - 1 <= 200010 && x - 1 >= 0 && book[x - 1] == 0) {			//判断边界，判断走没走，
			book[x - 1] = 1;
			q.push(x -1);
			step[x - 1] = step[x] + 1;
		}
	}
}
int main() {
	while (cin >> n >> k) {
		memset(step, 0, sizeof(step));				//不要忘记初始化 
		memset(book, 0, sizeof(book));
	if (n >= k)cout << n - k << endl;				//n不大于k的话  就是n-k了 
	else {
	bfs(n, k);
	cout << step[k] << endl;
	}
	}
	return 0;
}