poj 3278 Catch That Cow

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.

思路

入门bfs.这是一道一维空间上的bfs,比较简单,但多数情况是在二维甚至三维空间上bfs.此时需要用结构体保存不同坐标轴上的位置.

/*************************************************************************
> File Name: poj3278.cpp
> Author:ukiy
> Mail:
> Created Time: 2016年12月03日 星期六 19时03分45秒
************************************************************************/

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<string>
#include<cstring>
#include<vector>
#include<set>
#include<list>
#include<map>
#include<queue>
int dire[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
int dire2[8][2]={{-1,-1},{-1,0},{-1,1},{ 0,-1},{ 0,1},{ 1,-1},{ 1,0},{ 1,1}};
#define rep(i,a,b) for(int i=(a);i<=(b);(i++))
#define inf 0x3f3f3f
#define ll long long
#define pi acos(-1)
int dire3[6][3]={ {0,0,1},{0,1,0},{1,0,0},{0,0,-1},{0,-1,0},{-1,0,0} };
using namespace std;
const int  maxn=100001;
int visit[maxn+10];
int step[maxn+10];
queue<int> q;

int n,k;
int main()
{
std::ios::sync_with_stdio(false);
#ifndef OnlineJudge
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
#endif
cin>>n>>k;
memset(visit,0,sizeof(visit));
q.push(n);
step
=0;
visit
=1;
while(!q.empty()){
int  head,next;
head=q.front();
q.pop();
rep(i,0,2){
if(i==0)
next=head-1;
if(i==1)
next=head+1;
if(i==2)
next=head*2;
if(next>=0 &&next<=maxn &&!visit[next]){
q.push(next);
step[next]=step[head]+1;
visit[next]=1;
}

if(next==k){
printf("%d\n",step[next]);
return 0;
}
}
}

return 0;
}