poj 3278 -- Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 46279		Accepted: 14508

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 pointK (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.

---

思路：广搜的水题。

 

/*======================================================================
 *           Author :   kevin
 *         Filename :   CatchThatCow.cpp
 *       Creat time :   2014-08-03 10:06
 *      Description :
========================================================================*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#define clr(a,b) memset(a,b,sizeof(a))
#define M 200005
using namespace std;
int vis[M],cnt[M];

void BFS(int m,int n)
{
    queue<int>que;
    vis[m]=1;
    que.push(m);
    while(que.front()!=n && !que.empty()){
        int t=que.front();
        que.pop();
        if(!vis[t-1] && (t-1>=0 && t-1<M)){
            que.push(t-1);
            cnt[t-1]=cnt[t]+1;
            vis[t-1]=1;
        }
        if(!vis[t+1] && (t+1>=0 && t+1<M)){
            que.push(t+1);
            cnt[t+1]=cnt[t]+1;
            vis[t+1]=1;
        }
        if(!vis[t*2] && (t*2>=0 && t*2<M)){
            que.push(t*2);
            cnt[t*2]=cnt[t]+1;
            vis[t*2]=1;
        }
    }
}

int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF){
        clr(vis,0);
        clr(cnt,0);
        BFS(n,k);
        printf("%d\n",cnt[k]);
    }
    return 0;
}