bfs hdu 2717 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14553    Accepted Submission(s): 4422

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.

 

#include<iostream>
#include<queue>
#include<string.h>

using namespace std;

int k;
int vis[100005];
struct node
{
    int data;
    int step;
};
node n;
int bfs()
{
    queue<node>que;
    que.push(n);
    while(!que.empty())
    {
        node p = que.front();
        que.pop();
        vis[p.data]=1;
        if(p.data==k)
        {
            return p.step;
        }
        node q = p;
        q.step++;
        
        q.data = p.data*2;
        if(q.data>=0 && q.data<=100000 && !vis[q.data])
            que.push(q);
        q.data=p.data-1;
        if(q.data>=0 && q.data<=100000 && !vis[q.data])
            que.push(q);
        q.data=p.data+1;
        if(q.data>=0 && q.data<=100000 && !vis[q.data])
            que.push(q);
        
    }
    return -1;
}
int main()
{
    while(cin>>n.data>>k)
    {
        memset(vis,0,sizeof(vis));
        cout<<bfs()<<endl;
    }
    
    
    return 0;
}