bfs—Catch That Cow

bfs—Catch That Cow—poj3278

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 87152		Accepted: 27344

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

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

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.

简单的bfs，题目是单测试例，写的是多测试例。

 1 #include<iostream>
 2 #include<queue>
 3 #include<cstring>
 4 using namespace std;
 5 
 6 const int MAX=100000;
 7 int pace[MAX+10];
 8 int main()
 9 {
10     int n,k;
11     while(cin>>n>>k)
12     {
13         memset(pace,0,sizeof(pace));
14         int t;
15         queue<int> qu;
16         qu.push(n);
17         while(!qu.empty())
18         {
19             t=qu.front();
20             if(t==k)
21             {
22                 cout<<pace[t]<<endl;
23                 return 0;
24             }
25             qu.pop();
26             if(t<MAX&&!pace[t+1])
27             {
28                 qu.push(t+1);
29                 pace[t+1]=pace[t]+1;
30             }
31             if(t>0&&!pace[t-1])
32             {
33                 qu.push(t-1);
34                 pace[t-1]=pace[t]+1;
35             }
36             if(t*2<=MAX&&!pace[t*2])
37             {
38                  qu.push(t*2);
39                  pace[t*2]=pace[t]+1;
40             }
41         }
42     }
43     return 0;
44 }