Catch That Cow

Catch That Cow

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 113   Accepted Submission(s) : 46

Problem 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

 

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
const int MAXN=100010;
int visit[MAXN];
int now[3];
int N,K,step,t;
void bfs(){
    queue<int>dl;
    dl.push(N);
    visit[N]=1;
    if(N!=K){
    while(!dl.empty()){
        t=dl.size();
        step++;
        while(t--){
            now[0]=dl.front()+1;
            now[1]=dl.front()-1;
            now[2]=dl.front()*2;
            dl.pop();
            for(int i=0;i<3;i++){
                if(now[i]==K)return;
                if(now[i]>0&&now[i]<=100000&&!visit[now[i]]){//醉了，少了个0，错了n次；；；；； 
                    visit[now[i]]=1;dl.push(now[i]);
                }
            }
        }
    }
}
}
int main(){
    while(~scanf("%d%d",&N,&K)){
        memset(visit,0,sizeof(visit));
        step=0;
        bfs();
        printf("%d
",step);
    }
    return 0;
}