# bzoj1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

## 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?

农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．
他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有

那么，约翰需要多少时间抓住那只牛呢？

## Input

* Line 1: Two space-separated integers: N and K

仅有两个整数N和K.

## Output

* Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

最短的时间．

## Sample Input

5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.

## Sample Output

4

OUTPUT DETAILS:

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.

 1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<queue>
5 using namespace std;
6 #define N 100002
7 int vis[N<<1],s,t;
8 queue <int> h;
9 int main(){
10     scanf("%d%d",&s,&t); h.push(s);
11     if(s==t){printf("0");return 0;}
12     while(!h.empty()){
13         int x=h.front(); h.pop();
14         if(!vis[x-1]&&x) vis[x-1]=vis[x]+1,h.push(x-1);
15         if(!vis[x+1]&&x+1<N) vis[x+1]=vis[x]+1,h.push(x+1);
16         if(!vis[x<<1]&&(x<<1)<N) vis[x<<1]=vis[x]+1,h.push(x<<1);
17         if(vis[t]){printf("%d",vis[t]);return 0;}
18     }
19 }
View Code

posted @ 2018-11-28 14:00  kafuuchino  阅读(107)  评论(0编辑  收藏  举报