POJ 3278 Catch That Cow

POJ 3278 Catch That Cow

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.

This question is one of the simplest questions I solved or will solve.But not a very easy one. At the first time I used DFS instead of BFS,which is wrong.I never thought about what kind of questions can the two methods to be used.So I tried a lot of time in DFS,failed eventually. But after tried BFS,I finally understand the essence of BFS. It’s very happy and important for me to learn something new from the past knowledge.

So it’s BFS,I added all of the three situations to the queue.

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;
const int B=1000000;
int n,k;
int m[1000000+1];
struct nodes
{
int x,step;
};
int check(int x)
{
if(x<0||x>=B||m[x])
{
return 0;
}
else
{
return 1;
}
}
int bfs(int x)
{
queue<nodes> Q;
nodes a,next;
a.x=x;
a.step=0;
m[x]=1;
Q.push(a);
while(!Q.empty())
{
a=Q.front();
Q.pop();
if(a.x == k)
{
return a.step;
}
next=a;
next.x=a.x+1;
if(check(next.x))
{
next.step=a.step+1;
m[next.x]=1;
Q.push(next);
}
next.x=a.x-1;
if(check(next.x))
{

next.step=a.step+1;
m[next.x]=1;
Q.push(next);
}
next.x=a.x*2;
if(check(next.x))
{

next.step=a.step+1;
m[next.x]=1;
Q.push(next);
}

}
return -1;
}
int main()
{

int ans;
while(cin>>n>>k)
{
memset(m,0,sizeof(m));
ans=bfs(n);
cout<<ans<<endl;
}
return 0;
}