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Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 100270 Accepted: 31366
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.

题意:从x走到y点,每一步都有三种方案,走到2倍的位子或者加1或者减1。

思路:找的是最短的走法,是广搜的模版题。

复制代码
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#include<stdio.h> #include<iostream> #include<algorithm> #include<string.h> #include<queue> #include<string> using namespace std; struct node { int x,l; } nod[100010]; int n,k,ans; int vis[200010]; int bfs() { queue<node>q; while(!q.empty()) q.pop(); vis[n]=1; node t; t.x=n; t.l=0; q.push(t); while(!q.empty()) { t=q.front(); q.pop(); if(t.x==k) return t.l; for(int j=0; j<3; j++) { node T=t; if(j==0) { T.x*=2; T.l+=1; } if(j==1) { T.x+=1; T.l+=1; } if(j==2) { T.x-=1; T.l+=1; } if(T.x==k) return T.l; if(T.x<200005&&T.x>=0&&vis[T.x]==0) { vis[T.x]=1; q.push(T); } } } return 0; } int main() { while(~scanf("%d%d",&n,&k)) { memset(vis,0,sizeof(vis)); cout<<bfs()<<endl; } }



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