POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 90510 Accepted: 28386

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

USACO 2007 Open Silver

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <queue>#include <cctype>#define N 100010int Map[N*2];//判断数字有没有被访问int n,k;using namespace std;struct node{    int x;    int step;};int check(int x){    if(x<0||x>N||Map[x])        return 0;    return 1;}int bfs(int t){    queue<node> Q;    node a,next;//x为当前点    a.x=t;    a.step=0;    Map[t]=1;//表示当前点被访问过    Q.push(a);//把a入队    while(!Q.empty())    {        a=Q.front();        Q.pop();//读出当前队首元素        if(a.x==k)return a.step;        next=a;//下一个结点的可能        //每次都将三种状况加入队列之中        next.x=a.x+1;        if(check(next.x))        {            next.step=a.step+1;            Map[next.x]=1;            Q.push(next);        }        next.x=a.x-1;        if(check(next.x))        {            next.step=a.step+1;            Map[next.x]=1;            Q.push(next);        }        next.x=a.x*2;        if(check(next.x))        {            next.step=a.step+1;            Map[next.x]=1;            Q.push(next);        }    }    return 0;//表示没有找到}int main(){    while(cin>>n>>k)    {        memset(Map,0,sizeof(Map));        cout<<bfs(n)<<endl;    }    return 0;}