HDU 2717 Catch That Cow(bfs)

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 10575    Accepted Submission(s): 3303

 

 

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.

 

 

 

Source

USACO 2007 Open Silver

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
long long l,k;
const int maxn=100000;
bool v[100010];
struct  node
{
    long long x;
    int step;
};
bool check(long long a)
{
    if(a<0||a>maxn||v[a]) return false;
    return true;
}
int bfs()
{
    node n,m;
    n.x=l;
    n.step=0;
    queue<node> q;
    q.push(n);
    v[n.x]=1;
    while(!q.empty()){
        n=q.front();
        q.pop();
        if(n.x==k) return n.step;
        //-1
        m.x=n.x-1;
        if(check(m.x)){
            m.step=n.step+1;
            v[m.x]=1;
            q.push(m);
        }
        //+1
        m.x=n.x+1;
        if(check(m.x)){
            m.step=n.step+1;
            v[m.x]=1;
            q.push(m);
        }
        //*2
        m.x=n.x*2;
        if(check(m.x)){
            m.step=n.step+1;
            v[m.x]=1;
            q.push(m);
        }
    }
    return -1;
}
int main()
{
    while(cin>>l>>k){
        memset(v,0,sizeof(v));
        cout<<bfs()<<endl;
    }
    return 0;
}