Catch That Cow (BFS)POJ

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.

我的题解:这个是一维的BFS,和二维的差不多;不过临界值题目没告诉我们,应该要比1e5+1大才好.

#include <iostream>
#include <queue>
#include <cstring>
const int N=1e5+5;
using namespace std;
int a[N];
int k;
int bfs(int now){
    queue<int> q;
    q.push(now);
    while(!q.empty()){
        now=q.front();q.pop();
        if(now==k) return a[now];
        if(now+1<=N-4&&a[now+1]==-1) q.push(now+1),a[now+1]=a[now]+1;
        if(now-1>=0&&a[now-1]==-1) q.push(now-1),a[now-1]=a[now]+1;
        if(now*2<=N-4&&a[now*2]==-1) q.push(now*2),a[now*2]=a[now]+1;
    }
    return 0;
}
int main()
{
    int n;
    cin>>n>>k;
    memset(a,-1,sizeof(a));
    a[n]=0;
    cout<<bfs(n)<<endl;
    //cout << "Hello world!" << endl;
    return 0;
}