Catch That Cow

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.

提示：数组是1000000以上；

C++版本一

BFS

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <cstring>
#include <queue>
using namespace std;
int a[10000010];
long n,k;

struct node
{
     long x;
     long cnt;

}front1,tmp,start;
    queue <node> Q;

long bfs(node s){

    while(!Q.empty())
        Q.pop();

    if(s.x==k)return s.cnt;

	Q.push(s);
    while(!Q.empty()){
        front1=Q.front();
        Q.pop();

        for(int i=-1;i<=1;i++){
            if(i==0)
                tmp.x=2*front1.x;
            else
                tmp.x=front1.x+i;
            if(tmp.x>=0&&tmp.x<=1000000&&a[tmp.x]==0){
                a[tmp.x]=1;
                tmp.cnt=front1.cnt+1;
              //printf("%ld  \n",tmp.x);
                Q.push(tmp);
          	}
          	if(tmp.x==k) return tmp.cnt;
        }
    }
    return -1;
}

int main()
{
    while(~scanf("%ld%ld",&n,&k)){
        getchar();
            memset(a,0,sizeof(a));
            a[n]=1;
            //ans=0;
            start.x=n;
            start.cnt=0;

            printf("%ld\n",bfs(start));

    }

   //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

BFS

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
 
const int N = 1000000;
int map[N+10];
int n,k;
struct node
{
    int x,step;
};
 
int check(int x)
{
    if(x<0 || x>=N || map[x])
        return 0;
    return 1;
}
 
int bfs(int x)
{
    int i;
    queue<node> Q;
    node a,next;
    a.x = x;
    a.step = 0;
    map[x] = 1;
    Q.push(a);
    while(!Q.empty())
    {
        a = Q.front();
        Q.pop();
        if(a.x == k)
            return a.step;
        next = a;
        //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 -1;
}
 
int main()
{
    int ans;
    while(~scanf("%d%d",&n,&k))
    {
        memset(map,0,sizeof(map));
        ans = bfs(n);
        printf("%d\n",ans);
    }
    return 0;
}