题解 | #Catch That Cow#（BFS）

原题链接

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:  and 

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=100001;
struct status{//状态
  int n;
  int t;
  status(int x,int y):n(x),t(y){}
};

bool visit[maxn];//访问标记

 int bfs(int n,int k){
   queue<status>myque;
   myque.push(status(n,0));//初始状态入队
   visit[n]=true;
   
   while(!myque.empty()){//广度优先搜索
     status cur=myque.front();//取队头
     myque.pop();
     if(cur.n==k)return cur.t;//访问队头
     
     for(int i=0;i<3;i++){//状态转移
       status next(cur.n,cur.t+1);//时间++
       if(i==0)next.n+=1;
       else if(i==1)next.n-=1;
       else if(i=2)next.n*=2;
       
       if(next.n<0||next.n>maxn||visit[next.n])continue;//越界或者已被访问
       myque.push(next);
       visit[next.n]=true;
     }
   }
 }
 int main(){
   int n,k;
   while(scanf("%d%d",&n,&k)!=EOF){
     memset(visit,false,sizeof(visit));
     printf("%d\n",bfs(n,k));
   }
   return 0;
 }