最短パス問題の広さ優先検索練習

Catch That Cow
Time Limit: 2000MS
 
Memory Limit: 65536K
Total Submissions: 124627
 
Accepted: 38800
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.
 
人の位置は5で、牛の位置は7で、人は3種類の移動の方式X-1があって、X+1、2*X
1 D座標を頂点と見なし、到達可能な点を一歩移動して連通し、最短パスを探すのが広さ優先検索です.ここではすべての点へのパスを記録する必要があるので、1つの構造体で保存し、コードします.

#include
#include
#include
#define MAXN 100000
using namespace std;

int visited[MAXN+10];
int N, K;   //      N,     K
struct Step
{
	int x;       //   
	int steps;   //   x     
	Step(int xx,int s):x(xx),steps(s){}
};
void BFS();
queue q;

int main()
{
	cin >> N >> K;
	BFS();
}

void BFS()
{
	memset(visited, 0, sizeof(visited));  //           
	q.push(Step(N,0));                    //            
	visited[N] = 1;
	while (!q.empty())
	{
		Step s = q.front();
		if (s.x == K)
		{
			cout << s.steps << endl;       //         
			return;
		}
		else
		{
			if(s.x-1>=0 && !visited[s.x-1])
			{
				q.push(Step(s.x - 1, s.steps + 1));
				visited[s.x - 1] = 1;
			}

			if (s.x+1<=MAXN && !visited[s.x+1])
			{
				q.push(Step(s.x+1,s.steps+1)); 
				visited[s.x + 1] = 1;
			}

			if (s.x * 2 <= MAXN && !visited[s.x * 2])
			{
				q.push(Step(2 * s.x, s.steps + 1));
				visited[s.x * 2] = 1;
			}
		
			q.pop();
		}
		

	
	}
}