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;}