PoJ3278--Catch That Cow(Bfs)

Catch That Cow

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.

Source

USACO 2007 Open Silver

RE: 坐标轴上有两个点（a、b）, 求 a 接近 b 所需走的最小步数；

　　有三种走法，（x-1)、 （x+1）、（2*x）;

　　a > b  → → 输出；

　　else :: Bfs;

 1 #include <queue>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 using namespace std;
 6
 7 const int maxn = 100000 + 5;
 8 int vis[maxn], cnt[maxn];
 9 queue <int> q;
10
11 void Bfs()
12 {
13     int nx;
14     while(!q.empty())
15     {
16         int x = q.front();  q.pop();
17         for(int i = 0; i < 3; i++)    //三种走法；
18         {
19             if(i == 0) nx = x + 1;
20             else if(i == 1) nx = x - 1;
21             else nx = 2 * x;
22             if(nx >= 0 && nx <= 100000 && !vis[nx])  //遍历到最后找最优？::剪枝？
23             {
24                 vis[nx] = 1;
25                 cnt[nx] = cnt[x] + 1;
26                 q.push(nx);
27             }
28         }
29     }
30 }
31
32 int main()
33 {
34     int m, n;
35     while(~scanf("%d %d", &m, &n))
36     {
37         if(m > n)
38         {
39             printf("%d\n", m - n);
40             return 0;
41         }
42         memset(vis, 0, sizeof(vis));
43         vis[m] = 1;
44         cnt[m] = 0;
45         q.push(m);
46         Bfs();
47         printf("%d\n", cnt[n]);
48     }
49     return 0;
50 }