Catch That Cow(BFS)

Link：http://poj.org/problem?id=3278

problem：

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 47446		Accepted: 14895

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.

code：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
using namespace std;
#define maxn 100000
int step[100011],visit[100011];
queue<int>Q;
int bfs(int n,int k)
{
int now,next,i;
Q.push(n);
step[n]=0;
visit[n]=1;
while(!Q.empty())
{
now=Q.front();
Q.pop();
for(i=1;i<=3;i++)
{
if(i==1)
next=now-1;
else if(i==2)
next=now+1;
else
next=now*2;
if(next<0||next>maxn)
continue;
if(!visit[next])
{
visit[next]=1;
Q.push(next);
step[next]=step[now]+1;
}
if(next==k)
return step[next];
}
}
}
int main()
{
int n,k;
while(scanf("%d%d",&n,&k)!=EOF)
{
while(!Q.empty())
Q.pop();
memset(visit,0,sizeof(visit));
memset(step,0,sizeof(step));
printf("%d\n",bfs(n,k));
}
return 0;
}