杭电2717 Catch That Cow （bfs入门）

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19774    Accepted Submission(s): 5805

 

Problem 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.

 

 

Source

USACO 2007 Open Silver

 

dfs明显会超时，用bfs写在大佬郭聚聚指导下学会了bfs，可是用郭聚聚方法疯狂re，可能是oj针对我吧；

和聚聚聊天中突然想到可以不用结构体做，直接每次将两个元素穿进队列里面就ok

下面附上蒟蒻代码

#include 
#define ma 100010
typedef long long ll;
using namespace std;
int bfs(ll x,ll y)
{
    queueq;
    q.push(x);
    q.push(0);
    bool seen[ma] = {0};
    while(!q.empty())
    {
        ll num = q.front();
        q.pop();
        ll step = q.front();
        q.pop();
        if(num == y)
        {
            return step;
        }
        else
        {
            if(num - 1 >= 0 && !seen[num-1])
            {
                seen[num-1] = 1;
                q.push(num-1);
                q.push(step+1);
            }
            if(num * 2 <= ma && !seen[num*2])
            {
                seen[num*2] = 1;
                q.push(num*2);
                q.push(step+1);
            }
            if(num + 1 <= ma && !seen[num+1])
            {
                seen[num+1] = 1;
                q.push(num+1);
                q.push(step+1);
            }
        }
    }
}
int main()
{
    ll n,m;
    while(cin >> n >> m)
    {
        cout << bfs(n,m) << endl;
    }
    return 0;
}