A - 广搜 基础

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A - 广搜 基础

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3278

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.

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//@author:  yzj   Date:2015/07/26//sourse :  http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82603#problem/A//meaning:  农夫追牛，农夫每次可以走2*x, x-1 或 x+1三种方式，求最小时间#include <iostream>#include <cstdio>#include <queue>using namespace std;const int  MAXN  = 100000+5;int n, k;bool vis[MAXN];int step[MAXN];queue<int> que;int bfs(int pos){    que.push(pos);    vis[pos] = true;    step[pos] = 0;    while(!que.empty())    {        int head, next;        head = que.front();        que.pop();        for(int i = 0; i < 3; i++)//枚举三种方法        {            if(i == 0)            {                next = head * 2;            }            else if(i == 1)            {                next = head + 1;            }            else            {                next = head - 1;            }            //超出界限， 跳过            if(next < 0 || next > MAXN || vis[next]) continue;            else            {                que.push(next);                step[next] = step[head] + 1;                vis[next] = true;            }            //广搜的特性，返回的第一个一定是最短的路径            if(next == k) return step[next];        }    }}int main(){    //freopen("f:/yzj/cppCode/input.txt", "r", stdin);    scanf("%d %d", &n, &k);    if(n >= k) printf("%d\n", n-k);    else printf("%d\n", bfs(n));    return 0;}