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最新推荐文章于 2024-11-25 22:26:59 发布

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最新推荐文章于 2024-11-25 22:26:59 发布

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原文链接：http://www.cnblogs.com/ldxsuanfa/p/10953763.html

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<span style="color:#330099;">/*
A - 广搜 基础
Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit
 
Status
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
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.
By Grant Yuan
2014.7.13
*/
#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>

转载于:https://www.cnblogs.com/ldxsuanfa/p/10953763.html