Catch That Cow

Posted 2021-03-14 whitabbit

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.

 

给你两个数N和K,你可以对N进行三种操作,N=N-1,N=N+1,N=N*2,要求在操作数最小的情况下,使N==K,直接BFS暴力搜就行,可以分为以下三种情况进行剪枝(当然,暴力将100000个点搜完也能AC)

令N=N-1时,保证next>=0

令N=N+1时,保证next<=k

令N=N*2时,保证next<=k或者next-e+1<e-no

　　当next<=k,一定能减少到达终点时的操作数(当然,N==1时不影响总操作数);

　　当next>e,我们这时剩余部分的步数已经确定了(因为只能通过N=N-1使N==K),这时就要比较一下进行该操作和不进行该操作的情况下,e-now就时不进行操作的剩余步数,next-e就是进行操作时的剩余操作数,由于N=N*2也花费了一次操作所以应该加1,而两者相同的情况下自然也不必进行这个操作,毕竟操作数都是一样的

　　注意:总共只有100000个点,所以也要保证next<=100000

 

#include <algorithm>
#include <bitset>
//#include <bits/extc++.h>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>

using namespace std;
//using namespace __gnu_pbds

#define ll long long
#define maxn 105

int s, e, step[100005];
bool vis[100005];

void bfs()
{
    queue<int> q;
    q.push(s);
    vis[s] = 0;
    step[s] = 0;
    while (!q.empty())
    {
        int now = q.front();
        q.pop();
        int next = now + 1;
        if (next <= e && vis[next])
        {
            vis[next] = false;
            step[next] = step[now] + 1;
            q.push(next);
        }
        if (next == e)
        {
            return;
        }
        next = now - 1;
        if (next >= 0 && vis[next])
        {
            vis[next] = false;
            step[next] = step[now] + 1;
            q.push(next);
        }
        if (next == e)
        {
            return;
        }
        next = now << 1;
        if ((next <= e || next - e + 1 < e - now) && next <= 100000 && vis[next])
        {
            vis[next] = false;
            step[next] = step[now] + 1;
            q.push(next);
        }
        if (next == e)
        {
            return;
        }
    }
}

int main()
{
    while (~scanf("%d%d", &s, &e))
    {
        memset(vis, true, sizeof(vis));
        if (s >= e)
        {
            printf("%d
", s - e); //剪枝
        }
        else
        {
            bfs();
            printf("%d
", step[e]);
        }
    }
    return 0;
}