Catch That Cow（bfs）（注意边界条件）

Problem Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. 农民J 被告知 fugitive牛的定位，并且想立马逮到他！ 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. 他在一个点N上开始走，牛在K上， Farmer John has two modes of transportation: walking and teleporting. 交通：走/电运 * Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute 走路：一分钟走到 x-1 x+1 * Teleporting: FJ can move from any point X to the point 2 × X in a single minute. 电运：X走到 2X  一分钟 If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it? 多次时间抓到？   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.

#include <iostream>
#include <queue>
using namespace std;

typedef struct Now {
	int pos;
	int t;
} Now;

bool mark[100001];
queue<Now> q;

int bfs(int n,int k) {
	while(!q.empty()) {
		q.pop();
	}

	Now chu;
	chu.pos = n;
	chu.t = 0;

	Now father;
	Now child;

	q.push(chu);
	while(!q.empty()) {
		father = q.front();
		q.pop();


		if (father.pos + 1 >= 0 && father.pos + 1 <= 100000 && mark[father.pos + 1] == 0) {
			child = father;
			child.pos ++;
			child.t++;

			if (child.pos == k) {
				return child.t;
			}
			mark[child.pos] = 1;
			q.push(child);
		}

		if (father.pos - 1 >= 0 && father.pos - 1 <= 100000 && mark[father.pos - 1] == 0) {
			child = father;
			child.pos --;
			child.t++;
			if (child.pos == k) {
				return child.t;
			}
			mark[child.pos] = 1;
			q.push(child);
		}

		if (father.pos * 2 >= 0 && father.pos * 2 <= 100000 && mark[father.pos * 2] == 0) {
			child = father;
			child.pos *= 2;
			child.t++;
			if (child.pos == k) {
				return child.t;
			}
			mark[child.pos] = 1;
			q.push(child);
		}

	}

	return -1;

}


int main() {
	int n;
	int k;
	while (cin >> n >> k) {
		for (int i = 0; i <= 100000; i++) {
			mark[i] = 0;
		}
		if (n == k) {
			cout << 0 << endl;
		} else {

			int res = bfs(n,k);
			cout << res << endl;

		}

	}


}