poj--3278

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 65426 Accepted: 20554

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.

解体思路：简单的bfs。

代码如下：

#include<stdio.h>#include<queue>#include<string.h>using namespace std;int n,k,ans;struct stu{int x,t;};stu node[200000];int vist[200000];bool check(int x){if(vist[x]==1)return false;if(x<0)return false;if(x>=200000)return false;return true;}void bfs(int n){queue<stu >q;while(!q.empty()){q.pop();}stu temp,star;temp.x=n;temp.t=0;q.push(temp);while(!q.empty()){ star=q.front(); q.pop();  for(int i=1;i<=3;i++){  if(i==1){  temp.x=star.x+1;  temp.t=star.t+1;  }  else if(i==2){  temp.x=star.x-1;  temp.t=star.t+1;  }  else if(i==3){  temp.x=star.x*2;  temp.t=star.t+1;  }  if(check(temp.x)){  if(temp.x==k){  ans=temp.t;  return ;  }  q.push(temp);  vist[temp.x]=1;  }  }}}int main(){while(scanf("%d%d",&n,&k)!=EOF){memset(vist,0,sizeof(vist));vist[n]=1;bfs(n);printf("%d\n",ans);}return 0;}