# BZOJ1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

## 1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 634  Solved: 310
[Submit][Status]

## 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.

## Source

 1 #include<cstdio>
2 #include<cstdlib>
3 #include<cmath>
4 #include<cstring>
5 #include<algorithm>
6 #include<iostream>
7 #include<vector>
8 #include<map>
9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 200000+1000
14 #define maxm 400000
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 using namespace std;
20 {
21     int x=0,f=1;char ch=getchar();
22     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
23     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
24     return x*f;
25 }
26 struct edge{int go,next,w;}e[2*maxm];
28 bool v[maxn];
29 void ins(int x,int y)
30 {
32 }
33 void insert(int x,int y)
34 {
35     ins(x,y);ins(y,x);
36 }
37 void spfa()
38 {
39     for(int i=0;i<=n;++i) d[i]=inf;
40     memset(v,0,sizeof(v));
41     int l=0,r=1,x,y;q[1]=s;d[s]=0;
42     while(l!=r)
43     {
44         x=q[++l];if(l==maxn)l=0;v[x]=0;
46          if(d[x]+1<d[y=e[i].go])
47          {
48              d[y]=d[x]+1;
49              if(!v[y]){v[y]=1;q[++r]=y;if(r==maxn)r=0;}
50          }
51     }
52
53 }
54 int main()
55 {
56     freopen("input.txt","r",stdin);
57     freopen("output.txt","w",stdout);
66 }