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.
注意：//注意了！两个相关判断，vis数组范围注意。先判断em.n<110000才能看vis,否则越界
bfs模板，不是数学问题的，与每次多跳一格有这个技巧不同
#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;
int k;
struct node{
int n,t;
}s,em;
int vis[110005]={
0};//0未1有
int bfs(void){
queue<node>q;
s.t=0;
vis[s.n]=1;
q.push(s);
while(!q.empty()){
s=q.front();
q.pop();
if(s.n==k)return s.t;
em.t=s.t+1;//下一个
//x-1
em.n=s.n-1;
if(vis[em.n]==0&&em.n>=0){
vis[em.n]=1;
q.push(em);
}
if(s.n<k)
{
//x+1
em.n=s.n+1;
if(em.n<110000&&vis[em.n]==0){
//注意了！两个相关判断，vis数组范围注意。先判断em.n<110000才能看vis,否则越界
vis[em.n]=1;
q.push(em);
}
//2x
em.n=2*s.n;
if(em.n<110000&&vis[em.n]==0){
vis[em.n]=1;
q.push(em);
}
}
}
}
int main()
{
scanf("%d%d",&s.n,&k);
printf("%d",bfs());
return 0;
}