poj3278——Catch That Cow（BFS，剪枝）-白红宇

poj3278——Catch That Cow（BFS，剪枝）

阅读量：2344 次

发布时间：2019-05-10

本文共 1753 字，大约阅读时间需要 5 分钟。

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

n只有三种变换情况，+1，-1，x2，求n最少能经过多少步成为k。

乍看很简单，对三种情况BFS就行了，但这个数据很大，容易TLE或者RE。

我第一次TLE，之后就设置了一个观察数组，如果某次BFS时又碰到了这个数，这时的次数肯定不是最小的。

第二次RE，又考虑了n大于k的情况，这时候肯定只能一直-1，所以特判。又x2的操作容易超数组，所以如果某次x2超过了目标的两倍，那肯定是没用的，剪掉。到达负数也不行，剪掉。最后就过了

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            #include 
             
              #define MAXN 100010#define mod 100010#define INF 0x3f3f3f3fusing namespace std;int n,k,ans;int vis[MAXN*10];struct Node{ int num; int step;};void bfs(int n){ queue
              
                q; Node start,tmp1,tmp2; start.num=n; start.step=0; q.push(start); while(!q.empty()) { tmp1=q.front(); q.pop(); if(tmp1.num==k) { ans=min(ans,tmp1.step); return; } if(tmp1.num-1>=0) { tmp2.num=tmp1.num-1; tmp2.step=tmp1.step+1; if(!vis[tmp2.num]) { vis[tmp2.num]=1; q.push(tmp2); } } tmp2.num=tmp1.num+1; tmp2.step=tmp1.step+1; if(!vis[tmp2.num]) { vis[tmp2.num]=1; q.push(tmp2); } if(tmp1.num*2
               
                >n>>k) { ans=INF; if(n>=k) { cout<
                
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