POJ 3278 Catch That Cow

Posted 2021-02-09 fengzeng666

题目地址： https://vjudge.net/problem/POJ-3278

 

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 129558		Accepted: 40251

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.

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <queue>
 4 
 5 using namespace std;
 6 
 7 #define MAX 100000
 8 int visited[MAX+20];
 9 int lev[MAX+20];    //记录每个结点的层数
10 
11 void BFS(int N, int K)
12 {
13     memset(visited, 0, sizeof(visited));
14     memset(lev, 0, sizeof(lev));
15     lev[N] = 0;
16     visited[N] = 1;
17     queue<int>Q;
18     Q.push(N);
19 
20     while (!Q.empty())
21     {
22         int v = Q.front();
23         Q.pop();
24 
25         if (v == K)
26             cout << lev[v] << endl;
27 
28         if (v+1 <= MAX && !visited[v + 1])
29         {
30             lev[v + 1] = lev[v] + 1;
31             Q.push(v + 1);
32             visited[v + 1] = 1;
33         }
34         if (v-1 >= 0 && !visited[v - 1])
35         {
36             lev[v - 1] = lev[v] + 1;
37             Q.push(v - 1);
38             visited[v - 1] = 1;
39         }
40         if (v * 2 <= MAX && !visited[v * 2])
41         {
42             lev[v * 2] = lev[v] + 1;
43             Q.push(v * 2);
44             visited[v * 2] = 1;
45         }
46     }
47     
48 }
49 
50 int main()
51 {
52     int N, K;
53     while (cin >> N >> K)
54     {
55         BFS(N, K);
56     }
57     
58 
59     return 0;
60 }