bfs入门——Catch That Cow

原题：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 pointN (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 orX + 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 andK

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.

解题思路：一维上的bfs搜索，搜索的同时记录时间即可

代码实现：

#include<stdio.h>#include<queue>using namespace std;int bfs(int sx,int nx){    queue<int>q;    int dir[2]={-1,1};//一维方向向量    int use[110000]={0};//路径标记    int now,next;    now=sx;    q.push(now);//压入起点    while(q.size())    {        now=q.front();        for(int i=0;i<3;i++)        {            if(i==2)next=now*2;            else next=now+dir[i];            if(next>=0&&next<=100000&&!use[next])//判断是否越界或者搜过            {                use[next]=use[now]+1;                if(next==nx)break;//找到终点                q.push(next);            }        }         if(next==nx)break;//找到终点         else q.pop();    }    return use[next];}int main(){    int n,k;    scanf("%d%d",&n,&k);    if(n==k)printf("0");//若起点和终点重合，直接输出0，否则会输出2    else printf("%d",bfs(n,k));    return 0;}