Catch That Cow



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.

人在n处，牛在k处，人一次可以到加一，减一，或乘以2的位置上，求到牛的最短时间。感觉用了dp做，和搜索的思想类似吧，每次进队列，然后搜索+1，-1，乘2的位置，然后反复，一直到k处，跳出循环，注意每次判断要加上限定，不然会越界。

#include<cstdio>#include<cstring>#include<iostream>#include<queue>using namespace std;int dp[100005], leap[100005];queue<int>p;int main(){int i, j, m, n, ans, k,temp;cin >> temp >> k;memset(leap, 0, sizeof(leap));dp[k] = -1; dp[temp] = 0; leap[temp] = 1;p.push(temp);while (p.size()){n = p.front();if (!leap[n + 1]&&n+1<=100000&&n+1>=0){leap[n + 1] = 1;dp[n + 1] = dp[n] + 1;p.push(n + 1);}if (!leap[n - 1]&&n-1>=0&&n-1<=100000){leap[n - 1] = 1;dp[n - 1] = dp[n] + 1;p.push(n - 1);}if (n * 2 <= 100000&&n*2>=0){if (!leap[2 * n]){leap[2 * n] = 1;dp[2 * n] = dp[n] + 1;p.push(2 * n);}}p.pop();if (dp[k] != -1)break;}cout << dp[k] << endl;return 0;}