C - Catch That Cow POJ - 3278

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 <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <map>
#include <stack>
#include <queue>
#include <vector>
using namespace std;

typedef long long ll;
typedef struct node {
    int x, step;
} node;
const int maxn = 1e5 + 10;
int t, n, m, ans, cnt, sx, sy, sz, ex, ey, ez;
char mp[35][35][35];
int vis[maxn];
bool OutLimit(int x) {
    if(x < 0 || x > 100000) {
        return 1;
    }
    return 0;
}
int BFS() {
    queue<node> Q;
    node a, b;
    a.x = n;
    a.step = 0;
    vis[a.x] = 1;
    Q.push(a);
    while(!Q.empty()) {
        a = Q.front();
        Q.pop();
        if(a.x == m) {
            return a.step;
        }
        for(int i = 0; i < 3; i++) {
            if(i == 0) {
                b.x = a.x - 1;
            } else if(i == 2) {
                b.x = a.x + 1;
            } else {
                b.x = 2 * a.x;
            }
            b.step = a.step + 1;
            if(OutLimit(b.x)) {
                continue;
            }
            if(!vis[b.x]) {
                vis[b.x] = 1;
                Q.push(b);
            }
        }
    }
    return 0;
}
int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m;
    int ans = BFS();
    cout << ans << endl;
    return 0;
}