Problem 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.

思路

给定两个数，n，k，求从n到k所走的次数所少，每次可以加1，或减1或翻倍。

简单 bfs。

代码

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <iostream>
#include <queue>
using namespace std;

long int n, k;
int dir[2] = {-1, 1};
long int t;
long int now;
int visit[100010];
int step[100010];

int bfs()
{
    queue<long int> q;
    q.push(n);
    visit[n] = 1;

    while (!q.empty())
    {
        t = q.front();
        q.pop();
        if (t == k)
            return 0;
        for (int i = 0; i < 3; i++)
        {
            if (i == 2)
            {
                now = t * 2;
            }
            else
            {
                now = t + dir[i];
            }

            if (now < 0 || now > 100000)
                continue;
            if (!visit[now])
            {
                q.push(now);
                visit[now] = 1;
                step[now] = step[t] + 1;
            }
            if (now == k)
                return step[now];
        }
    }
    return -1;
}

int main()
{

    while (scanf("%ld%ld", &n, &k) != EOF)
    {
        memset(visit, 0, sizeof(visit));
        memset(step, 0, sizeof(step));
        printf("%d\n", bfs());
    }
    return 0;
}