POJ 3278 Catch That Cow

链接：https://vjudge.net/problem/POJ-3278

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.

 

#include <iostream>
#include <queue>
#include <cstring>
#include <cstdio>
using namespace std;
const int N = 1e5;
int n,k;
int walk[N + 10];
int main()
{
    scanf("%d%d",&n,&k);
    queue<int> q;
    q.push(n);//将起点加入队列
    while(!q.empty())
    {
        int now = q.front();
        q.pop();
        if(now == k)//找到就跳出
        {
            printf("%d
",walk[k]);
            break;
        }
        //三种状态：+1,-1,*2
        if(now - 1 >= 0 && !walk[now - 1])
        {
            q.push(now - 1);
            walk[now - 1] = walk[now] + 1;
        }
        if(now + 1 <= N && !walk[now + 1])
        {
            q.push(now + 1);
            walk[now + 1] = walk[ now ] + 1;
        }
        if(now*2 <= N && !walk[now*2])
        {
            q.push(now*2);
            walk[now*2] = walk[now] + 1;
        }
    }
    return 0;
}