Problem 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.
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
4HintThe 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.
BFS 注意n=k
//第一次写，木有经验，将就将就~~~~~
#include<iostream>
using namespace std;
#include<stdio.h>
#include<string.h>
#define MAX 100002
int queue[2*MAX];
int step[2*MAX];
int judge(int y)
{
if(y>=2*MAX||y<0||step[y]!=0) return 0;
else return 1;//step[y]!=0表示已经访问过
}
int bfs(int n,int k)
{
int x,front=0,rear=0;
if(n==k) return 0;
queue[front++]=n;
while(rear<front)
{
x=queue[rear++];//x 表示当前访问点
if(x+1==k||x-1==k||2*x==k) return step[x]+1;
if(judge(x-1)) {step[x-1]=step[x]+1; queue[front++]=x-1;}
if(judge(x+1)) {step[x+1]=step[x]+1; queue[front++]=x+1;}
if(judge(x*2)) {step[x*2]=step[x]+1; queue[front++]=x*2;}
}
return 0;
}
int main()
{
int n,k;
while(~scanf("%d%d",&n,&k))
{
memset(step,0,sizeof(step));
printf("%d\n",bfs(n,k));
}
return 0;
}