A广搜

<span style="color:#330099;">/*A - 广搜 基础Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64uSubmit StatusDescriptionFarmer 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?InputLine 1: Two space-separated integers: N and KOutputLine 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.Sample Input5 17Sample Output4HintThe 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.By Grant Yuan2014.7.13*/#include<iostream>#include<stdio.h>#include<string.h>#include<queue>#include<cstdio>using namespace std;bool a[100003];typedef struct{   int num;   int sum;}dd;queue<dd>q;int n,k;int res;void slove(){int m,count;   int x;   dd init;   while(!q.empty()){       if(q.front().num==n)       {  res=q.front().sum;           break;       }    m=q.front().num;    count=q.front().sum;   x=m-1;   if(a[x]==0)     {   init.num=x;         init.sum=count+1;         q.push(init);         a[x]=1;     }    x=m+1;     if(a[x]==0)     {        init.num=x;         init.sum=count+1;         q.push(init);         a[x]=1;     }     if(m%2==0){         x=m/2;      if(a[x]==0)     {         init.num=x;         init.sum=count+1;         q.push(init);         a[x]=1;     }}     q.pop();}}int main(){       scanf("%d%d",&n,&k);    memset(a,0,sizeof(a));    dd init;    init.num=k;    init.sum=0;    a[k]=1;    q.push(init);    slove();    cout<<res<<endl;    return 0;}</span>