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

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Output

Sample Input

5 17

Sample Output

4

Hint

Source

#include <iostream>
#include <cstdio>
#include <queue>

using namespace std;

queue <int> que;
int times[100001];//記錄花費的時間
int visited[100001];//標記數組，比如x*2已經到達的地方標記下來，x+1到達時就不需要走這了，因為x*2更快
int bfs(int n,int k);

int main()
{
    int N,K;//N：FJ,K:cow
    scanf("%d%d",&N,&K);
    que.push(N);
    times[N]=0;
    visited[N]=1;
    printf("%d\n",bfs(N,K));
    return 0;
}
int bfs(int n,int k){

    while(!que.empty()){
        int t1=que.front();
        que.pop();
        int t;
        for(int i=0;i<3;i++){//遍歷三種方式，相當於在圖中遍歷四個方向
            t=t1;
            if(i==0){
                t=t+1;
            }
            else if(i==1){
                t=t-1;
            }
            else{
                t=t*2;
            }
            //因為N的範圍是0~100000所以可能越界
            ///剛開始我沒考慮到，就runtime error了。。
            if(t<0||t>100001)
                continue;

            if(visited[t]==0){
                que.push(t);
                visited[t]=1;
                times[t]=times[t1]+1;
            }

            if(t==k){
                return times[t];
            }
        }
    }
}

參考自：http://blog.csdn.net/freezhanacmore/article/details/8168265

【BFS】Catch That Cow(POJ3278)