POJ-3278 Catch That Cow

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

---

 

大数组可以用vector

大数组可以用vector

大数组可以用vector

 

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

int N, K;
vector<int> data;

int main(void)
{
    int bfs(void);
    
    while(scanf("%d%d", &N, &K) != EOF)
    {
        for(int i = 0; i <= 2*K; i++)
            data.push_back(-1);
        
        if(K <= N)
        {
            printf("%d\n", N-K);
            continue;
        }
        else
        {
            int e = bfs();
            printf("%d\n", e);
        }
        
        
    }
    
    return 0;
}


int bfs()
{
    queue<int> que;
    int cas;
    
    data[N] = 0;
    cas = N;
    que.push(cas);
    
    while(que.size())
    {
        int ca = que.front();   que.pop();
        
        if(ca == K)
            break;
        
        for(int i = 0; i < 3; i++)
        {
            if(i == 0)
                cas = ca + 1;
            if(i == 1)
                cas = ca - 1;
            if(i==2)
                cas = ca * 2;
                
            if(cas >= 0 && cas <= 2*K && data[cas] < 0)
            {
                data[cas] = data[ca] + 1;
                que.push(cas);
            }
            
        }
        
        
    }
    
    
    return data[K];
}