POJ 3278(bfs)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 31637		Accepted: 9740

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.

//注意下标不可变为-1，两个数组均要开到20w，为此吃了好几个re 
//大致题意：n可以执行加一减一乘二，求至少几步到k 
#include <iostream>
#include <queue>
#include <cstring>
#include <cstdlib>
using namespace std;
const int N = 100000;
int a[200005];
queue <int > q;
int vis[N*2+5];
void bfs(int n,int k)
{
    vis[n] = true;
    q.push(n);
    while(!q.empty())
    {
        int head = q.front();
        q.pop();
        if(head==k)
        {
            return ;
        }
        else
        {//原来没加大于0的条件，结果re 
            if(!vis[head-1]&&head<=N&&head>=1)//禁止下标变为-1 
            {
                vis[head-1] = true;
                a[head-1]+=a[head]+1;
                q.push(head-1);
            }
            if(!vis[head+1]&&head<N&&head>=0)
            {
                vis[head+1] = true;
                a[head+1]+=a[head]+1;
                q.push(head+1);
            }
            if(!vis[head*2]&&head<=50000&&head>=0)
            {
                vis[head*2] = true;
                a[head*2]+=a[head]+1;
                q.push(head*2);
            }
        }
    }
}   
int main()
{
    int n,k;
    cin>>n>>k;
    memset(vis,false,sizeof(vis));
    memset(a,0,sizeof(a));
    while(!q.empty())
        q.pop();
    bfs(n,k);
    if(n>=k)
        cout<<n-k<<endl;
    else
        cout<<a[k]<<endl;
    system("pause");
    return 0;
}