Catch That Cow 分类： POJ 2015-06-29 19:06 10人阅读 评论(0) 收藏

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 58072		Accepted: 18061

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
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#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;

const int Max=1100000;
struct Point
{
    int x;
    int num;
};
int N,K;
bool vis[210000];
void BFS()
{
    queue<Point>a;
    Point s,t;
    memset(vis,false,sizeof(vis));
    s.num=0;
    s.x=N;
    a.push(s);
    vis[N]=true;
    while(!a.empty())
    {
        s=a.front();
        a.pop();
        if(s.x==K)
        {
            printf("%d\n",s.num);
            return ;
        }
        if(!vis[s.x+1]&&s.x+1<=K*2)
        {
            t.x=s.x+1;
            t.num=s.num+1;
            a.push(t);
            vis[t.x]=true;
        }
        if(s.x-1>=0&&!vis[s.x-1])
        {
            t.x=s.x-1;
            t.num=s.num+1;
            a.push(t);
            vis[t.x]=true;
        }
        if(s.x*2<=K*2&&!vis[s.x*2])
        {
            t.x=s.x*2;
            t.num=s.num+1;
            a.push(t);
            vis[t.x]=true;
        }
    }

}
int main()
{

    scanf("%d %d",&N,&K);
    BFS();
    return 0;
}

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