题目传送门
1 /*
2 BFS简单题：考虑x-1，x+1，x*2三种情况，bfs队列练练手
3 */
4 #include <cstdio>
5 #include <iostream>
6 #include <algorithm>
7 #include <map>
8 #include <queue>
9 #include <set>
10 #include <cmath>
11 #include <cstring>
12 using namespace std;
13
14 const int MAXN = 1e6 + 10;
15 const int INF = 0x3f3f3f3f;
16 int n, m;
17 int d[MAXN];
18 bool vis[MAXN];
19
20 void BFS(void)
21 {
22 memset (vis, 0, sizeof (vis));
23 for (int i=0; i<=1e6; ++i) d[MAXN] = 0;
24
25 queue<int> q;
26 q.push (n); d[n] = 0; vis[n] = true;
27
28 while (!q.empty ())
29 {
30 int x = q.front (); q.pop ();
31 if (x == m) break;
32
33 int xl = x - 1; int xr = x + 1; int x2 = x * 2;
34
35 if (xl >= 0 && !vis[xl])
36 {
37 q.push (xl); d[xl] = d[x] + 1;
38 vis[xl] = true;
39 }
40 if (xr <= 1e6 && !vis[xr])
41 {
42 q.push (xr); d[xr] = d[x] + 1;
43 vis[xr] = true;
44 }
45 if (x2 <= 1e6 && !vis[x2])
46 {
47 q.push (x2); d[x2] = d[x] + 1;
48 vis[x2] = true;
49 }
50 }
51
52 }
53
54 int main(void) //POJ 3278 Catch That Cow
55 {
56 //freopen ("POJ_3278.in", "r", stdin);
57
58 while (scanf ("%d%d", &n, &m) == 2)
59 {
60 BFS ();
61 printf ("%d
", d[m]);
62 }
63
64 return 0;
65 }