Catch That Caw

Catch That Caw

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.

本题寻找最短的步数，所以可以用BFS算法实现，每次对各个操作进行一次搜索，并存下当前位置

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
const int MAXN = 200005;
int vis[MAXN];//标记有没有达到过
int dx[] = { 0,0,1,-1 };
int dy[] = { 1,-1,0,0 };
const int INF = 1e9;
int  n, k;
struct node {int x, step;
};
queue<node> q;
void bfs() {int X, Step;while (!q.empty()) {node ND2 = q.front();q.pop();X = ND2.x;Step = ND2.step;if (X == k) {cout << Step << endl;return;}if (X >= 1 && vis[X - 1] == 0) {node ND3;vis[X - 1] = 1;ND3.x = X - 1;ND3.step = Step + 1;q.push(ND3);}if (X <= k && vis[X + 1] == 0) {node ND3;vis[X + 1] = 1;ND3.x = X + 1;ND3.step = Step + 1;q.push(ND3);}if (X <= k && vis[X * 2] == 0) {node ND3;vis[X * 2] = 1;ND3.x = X * 2;ND3.step = Step + 1;q.push(ND3);}}
}
void solve() {while (~scanf("%d%d", &n, &k)) {while (!q.empty())q.pop();//清空memset(vis, 0, sizeof vis);vis[n] = 1;//标记初始位置nnode ND1;ND1.x = n;ND1.step = 0;q.push(ND1);bfs();}
}int main() {//int _;//cin >> _;//while (_--) {solve();//{return 0;
}