POJ 3278 Catch That Cow

Posted 2020-12-03 ducklu

题意:给一个起点和终点，只有三种走法n+1, n-1, n*2, 求最短需要的步数

解题思路:BFS，用BFS解决无权图的最短路问题。

注:很容易re，我re了10多次，有两个点要注意

1.如果起始点s>=终点e那么，就只能回退，所以step=s-e

2.编号不可能出边界(0, MAXSIZE)

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <queue>
 5 #define MAXVERTEXNUM 1000100
 6 using namespace std;
 7 
 8 int dist[MAXVERTEXNUM];
 9 int Visited[MAXVERTEXNUM];
10 int s, e;
11 
12 void BFS()
13 {
14     queue<int> q;
15     q.push(s);
16     Visited[s] = 1;
17 
18     while (!q.empty())
19     {
20         int num = q.front();
21         if (num == e)
22             return;
23         q.pop();
24 
25         int d1 = num * 2, d2 = num + 1, d3 = num - 1;
26         if (d1 < MAXVERTEXNUM && !Visited[d1])
27         {
28             dist[d1] = dist[num] + 1;
29             q.push(d1);
30             Visited[d1] = 1;
31         }
32         if (!Visited[d2])
33         {
34             dist[d2] = dist[num] + 1;
35             q.push(d2);
36             Visited[d2] = 1;
37         }
38         if (d3 >= 0 && !Visited[d3])
39         {
40             dist[d3] = dist[num] + 1;
41             q.push(d3);
42             Visited[d3] = 1;
43         }
44     }
45 }
46 
47 int main()
48 {
49     cin >> s >> e;
50     memset(dist, 0, sizeof(dist));
51     memset(Visited, 0, sizeof(Visited));
52 
53     if (s >= e)
54         cout << s - e << endl;
55     else
56     {
57         BFS();
58         cout << dist[e] << endl;
59     }
60 
61     return 0;
62 }