[Poj 3278] Catch That Cow BFS

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 69092 Accepted: 21729

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.

三种情况的ｂｆｓ建边；

#include<iostream>
#include<queue>
#include<stdio.h>
#include<string.h>
using namespace std;
int vis[200030];
struct node{
int x,step;
node(int xx,int st)
{
x=xx;
step=st;
}node(){}
};
queue<node> q;
int ans;
int x,k;
int flag=0;
void bfs()
{
q.push(node(x,0));
vis[x]=1;
while(!q.empty())
{
node now=q.front();
q.pop();
if(flag==1) break;
int mx=x*2;
if(now.x>0&&vis[now.x-1]==0)
{
if(now.x-1==k)
{
flag=1;
ans=now.step+1;
}
vis[now.x-1]=1;
q.push(node(now.x-1,now.step+1));
}
if(now.x<k&&vis[now.x+1]==0)
{
if(now.x+1==k)
{
flag=1;
ans=now.step+1;
}
vis[now.x+1]=1;
q.push(node(now.x+1,now.step+1));
}
if(now.x<k&&vis[now.x*2]==0)
{
if(now.x*2==k)
{
flag=1;
ans=now.step+1;
}
vis[now.x*2]=1;
q.push(node(now.x*2,now.step+1));
}
}
}

int main()
{
scanf("%d%d",&x,&k);
bfs();
printf("%d\n",ans);

}