Poj 3278 Catch That Cow

1.Link:

http://poj.org/problem?id=3278

2.Content:

Catch That Cow
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 48245   Accepted: 15114

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 - 1 or + 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

3.Method:

4.Code:

 1 #include<iostream>
 2 #include<queue>
 3 
 4 using namespace std;
 5 
 6 //(0 ≤ N ≤ 100,000)
 7 #define MAX 200002
 8 int a[MAX];
 9 int main()
10 {
11     int i,j;
12     int x,y;
13     int m,n;
14     cin>>x>>y;
15     queue<int> q;
16     a[x]=1;
17     q.push(x);
18     while(!q.empty())
19     {
20         m=q.front();
21         if(m==y) {cout<<a[m]-1;break;}
22         else
23         {
24             n=m-1;
25             if(n>=0&&n<MAX&&a[n]==0)
26             {
27                 a[n]=a[m]+1;
28                 q.push(n);
29             }
30             n=m+1;
31             if(n>=0&&n<MAX&&a[n]==0)
32             {
33                a[n]=a[m]+1;
34                q.push(n);
35             }
36             n=m*2;
37             if(n>=0&&n<MAX&&a[n]==0)
38             {
39                a[n]=a[m]+1;
40                q.push(n);
41             }
42         }
43         q.pop();
44     }
45     //system("pause");
46     return 0;
47 }

5.Reference:

原文地址:https://www.cnblogs.com/mobileliker/p/4068288.html