题目

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.

题解

由于移动方式只有 +1 -1 *2 因此如果 起点大于终点，直接相减即可

剩下就是标准的BFS模板

代码

/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o<n;o++)

const int maxn = 100005;

int BFS(int s,int v) {
    if(s == v)
        return 0;
    if(s > v)
        return s - v;

    queue<int> Q;
    bool visited[maxn];
    memset(visited,false,sizeof(visited));
    int dis[maxn];
    memset(dis,0,sizeof(dis));

    Q.push(s);
    visited[s] = true;
    while(!Q.empty()) {
        int th = Q.front();
        Q.pop();

        //达到终点
        if(th == v)
            break;

        //拓展节点

       define push \
        if(next > maxn || next <= 0) \
            continue;\
        if(!visited[next]) {\
            Q.push(next);\
            visited[next] = true;\
            dis[next] = dis[th] + 1;\
        }\


        int next;
        next = th + 1;
        push;
        next = th - 1;
        push;
        next = th * 2;
        push;
    }

    if(dis[v])
        return dis[v];
    else
        return -1;
}

bool Do() {
    int s,v;
    if(scanf("%d%d",&s,&v)==EOF)
        return false;
    printf("%d\n",BFS(s,v));
    return true;
}

int main() {
    while(Do());
    return 0;
}