BFS POJ 3278 Catch That Cow

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\n", d[m]);
62     }
63
64     return 0;
65 }

转载于:https://www.cnblogs.com/Running-Time/p/4444809.html

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