POJ 3278 Catch That Cow

POJ: https://i.cnblogs.com/EditPosts.aspx?opt=1

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 80291		Accepted: 25297

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.

#include <iostream>
#include <queue>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;

const int N = 100009;

int main() {
    int visited[N];
    int n, k;
    while (cin>>n>>k) {
        queue<int>q;
        memset(visited, 0, sizeof(visited));
        q.push(n);
        while (!q.empty()) {
            int tmp = q.front();
            q.pop();
            if (tmp == k)
                break;
            if (tmp-1 >=0 && visited[tmp - 1] == 0) {
                visited[tmp - 1] = visited[tmp] + 1;
                q.push(tmp - 1);
            }
            if (tmp+1 <= k&&visited[tmp + 1] == 0) {
                visited[tmp + 1] = visited[tmp] + 1;
                q.push(tmp + 1);
            }
            if (tmp * 2 < N&&visited[2 * tmp] == 0) {
                visited[2 * tmp] = visited[tmp] + 1;
                q.push(2 * tmp);
            }
        }
        cout << visited[k] << endl;
    }
    return 0;

}