BFS

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.

 

题意 ： 农民有3种走路方式 ， 问怎么走可以最短到达目标位置。

 

思路 ： BFS 走感觉还是很好想到的 ， 之前一直有一个地方不是很理解 ，就是搜素图的时候什么要标记走过的点什么时候不标记走过的点，现在差不多懂点了 ，每走一步，就把当前这步所有可以到达的位置全部找到，之前走过的点不计入。还有在求步数的搜索题中，特意建一个数组，当前位置的步数等于上一次所走的步数 +1 。

 

代码 ：

/*
 * Author:  ry 
 * Created Time:  2017/10/26 9:33:23
 * File Name: 2.cpp
 */
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <time.h>
using namespace std;
const int eps = 1e5+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define ll long long
int n, k;
int bu[eps];

void bfs(){
    queue<int>que;
    que.push(n);
    //memset(vis, 0, sizeof(vis));
    memset(bu, -1, sizeof(bu));
    
    bu[n] = 0;
    while (!que.empty()){
        int a = que.front();
        que.pop();
        
        if (a == k) break;
        for(int i = 0; i < 3; i++){
            if (i == 0) {
                int b = a - 1;
                if (b >= 0 && b < eps && bu[b] == -1){
                    que.push(b);
                    bu[b] = bu[a] + 1;
                }
            }
            else if (i == 1){
                int b = a + 1;
                if (b >= 0 && b < eps && bu[b] == -1){
                    que.push(b);
                    bu[b] = bu[a] + 1;
                }
            }
            else {
                int b = 2*a;
                if (b >= 0 && b < eps && bu[b] == -1){
                    que.push(b);
                    bu[b] = bu[a] + 1;
                }
            }
        }        
    }    

}

int main (){
    
    while (~scanf("%d%d", &n, &k)){
        bfs();
        //for(int i = 1; i <= k; i++){
            //printf("%d	", bu[i]);
        //}
        //printf("
");
        printf("%d
", bu[k]);
    }
}