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

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.

 

简单的bfs，题目是单测试例，写的是多测试例。

 #include<iostream>
 #include<queue>
 #include<cstring>
 using namespace std;

 const int MAX=;
 int pace[MAX+];
 int main()
 {
     int n,k;
     while(cin>>n>>k)
     {
         memset(pace,,sizeof(pace));
         int t;
         queue<int> qu;
         qu.push(n);
         while(!qu.empty())
         {
             t=qu.front();
             if(t==k)
             {
                 cout<<pace[t]<<endl;
                 return ;
             }
             qu.pop();
             if(t<MAX&&!pace[t+])
             {
                 qu.push(t+);
                 pace[t+]=pace[t]+;
             }
             if(t>&&!pace[t-])
             {
                 qu.push(t-);
                 pace[t-]=pace[t]+;
             }
             if(t*<=MAX&&!pace[t*])
             {
                  qu.push(t*);
                  pace[t*]=pace[t]+;
             }
         }
     }
     return ;
 }