poj 3278 -- Catch That Cow

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概述

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 46279		Accepted: 14508

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 pointK (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

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.

思路：广搜的水题。

 1 /*======================================================================
 2  *
Author :
kevin
 3  *
Filename :
CatchThatCow.cpp
 4  *
Creat time :
2014-08-03 10:06
 5  *
Description :
 6 ========================================================================*/
 7 #include <iostream>
 8 #include <algorithm>
 9 #include <cstdio>
10 #include <cstring>
11 #include <queue>
12 #include <cmath>
13 #define clr(a,b) memset(a,b,sizeof(a))
14 #define M 200005
15 using namespace std;
16 int vis[M],cnt[M];
17
18 void BFS(int m,int n)
19 {
20
queue<int>que;
21
vis[m]=1;
22 
que.push(m);
23
while(que.front()!=n && !que.empty()){
24
int t=que.front();
25 
que.pop();
26
if(!vis[t-1] && (t-1>=0 && t-1<M)){
27
que.push(t-1);
28
cnt[t-1]=cnt[t]+1;
29
vis[t-1]=1;
30 
}
31
if(!vis[t+1] && (t+1>=0 && t+1<M)){
32
que.push(t+1);
33
cnt[t+1]=cnt[t]+1;
34
vis[t+1]=1;
35 
}
36
if(!vis[t*2] && (t*2>=0 && t*2<M)){
37
que.push(t*2);
38
cnt[t*2]=cnt[t]+1;
39
vis[t*2]=1;
40 
}
41 
}
42 }
43
44 int main()
45 {
46
int n,k;
47
while(scanf("%d%d",&n,&k)!=EOF){
48
clr(vis,0);
49
clr(cnt,0);
50 
BFS(n,k);
51
printf("%dn",cnt[k]);
52 
}
53
return 0;
54 }