poj--3278--Catch That Cow(bfs)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 67914		Accepted: 21397

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<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
#define MAXN 1000000+10
int vis[MAXN];
int ans,n,k;
struct node
{
int x,step;
}p,temp;
int check(int x)
{
if(x<0||x>=MAXN||vis[x])
return 0;
return 1;
}
int bfs()
{
queue<node>q;
p.step=0;
vis
=1;
p.x=n;
while(!q.empty()) q.pop();
q.push(p);
while(!q.empty())
{
p=q.front();
q.pop();
temp=p;
if(p.x==k)
return p.step;
temp.x=p.x+1;
if(check(temp.x))
{
vis[temp.x]=1;
temp.step=p.step+1;
q.push(temp);
}
temp.x=p.x-1;
if(check(temp.x))
{
vis[temp.x]=1;
temp.step=p.step+1;
q.push(temp);
}
temp.x=p.x*2;
if(check(temp.x))
{
vis[temp.x]=1;
temp.step=p.step+1;
q.push(temp);
}
}
return -1;
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
memset(vis,0,sizeof(vis));
ans=bfs();
printf("%d\n",ans);
}
return 0;
}