POJ 3278 Catch That Cow

Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 44070 Accepted: 13764

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

簡單的BFS應用，POJ刷題正式開始，poj計劃的開始就意味著，我以後的生活就沒以前這麼輕松了。

送自己一句話吧：想玩好ACM，8個字：“題海戰術！題海戰術！”

#include 
#include 
#include 
#include 
#include 
#include 
const int N =  10000010;
using namespace std;
int n,k;
struct node{
    int wz;
    int ans;
}q[N];
int vis[N];
int jz[] = {1,-1};
void BFS()
{
    int s = 0,e = 0;

    node f,t;
    f.ans = 0;
    f.wz = n;
    q[e++] = f;
    vis[n] = 1;
    while(s=0 && f.wz <=100000)
            {
                f.ans = t.ans + 1;
                vis[f.wz] = 1;
                q[e++] = f;
            }

        }
    }
}
int main()
{

    while(~scanf("%d%d",&n,&k))
    {
        memset(vis,0,sizeof(vis));
       BFS();
    }
    return 0;
}