Catch That Cow

Catch That Cow

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 3278

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.

简单题，不说，直接代码。。。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define MAX 100005
#define INF 0x3f3f3f3f
#define LL long long
#define pii pair<int,int>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rd3(x,y,z) scanf("%d%d%d",&x,&y,&z)
///map<int,int>mmap;
///map<int,int >::iterator it;
using namespace std;

struct Pos
{
    int x;
    int step;
    Pos() {}
    Pos(int x,int step)
    {
        this->x=x,this->step=step;
    }
};
bool vis[MAX];
int main ()
{
    int m,n;
    while(~rd2(m,n))
    {
        memset(vis,0,sizeof(vis));
        if(m>n)
        {
            printf("%d\n",m-n);
            continue;
        }
        queue<Pos> que ;
        que.push(Pos(m,0));
        while(!que.empty())
        {
            Pos temp=que.front();
            if(temp.x==n)
            {
                printf("%d\n",temp.step);
                break;
            }
            que.pop();
            if( temp.x-1>=0 && !vis[temp.x-1] )
            {
                que.push(Pos(temp.x-1,temp.step+1));
                vis[temp.x-1]=true;
            }
            if( temp.x+1<=MAX-5 && !vis[temp.x+1] )
            {
                que.push(Pos(temp.x+1,temp.step+1));
                vis[temp.x+1]=true;
            }
            if( temp.x<<1<=MAX-5 && !vis[temp.x<<1])
            {
                que.push(Pos(temp.x*2,temp.step+1));
                vis[temp.x<<1]=true;
            }
        }
    }
    return 0;
}