Day2-E-Catch That Cow-POJ3278

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.

 

分析：找最短路问题，首先想到BFS，三个状态分别为x-1，x+1，2*x，注意剪枝一些不太可能的状态，代码如下：

const int maxm = 1000100;

struct Node {
    int times, x;
    Node(int _times, int _x):times(_times),x(_x) {}
};

int vis[maxm], n, k;

int main() {
    scanf("%d%d", &n, &k);
    queue<Node> q;
    q.push(Node(0, n));
    while(!q.empty()) {
        Node tmp = q.front();
        q.pop();
        if(vis[tmp.x])
            continue;
        vis[tmp.x] = 1;
        if(tmp.x == k) {
            printf("%d
", tmp.times);
            break;
        }
        tmp.times++;
        if(tmp.x && !vis[tmp.x-1])
            q.push(Node(tmp.times, tmp.x - 1));
        if(!vis[tmp.x+1])
            q.push(Node(tmp.times, tmp.x + 1));
        if(tmp.x < k * 4 && !vis[tmp.x*2])
            q.push(Node(tmp.times, tmp.x * 2));
    }
    return 0;
}