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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 54696 | Accepted: 17101 |
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
Output
Sample Input
5 17
Sample Output
4
Hint
#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#define N 100009
using namespace std;
int n,k;
int vis[N];
int a[N];
int t;
int main()
{
while(~scanf("%d %d",&n,&k))
{
if(n==k)
{
cout<<0<<endl;
continue;
}
memset(a,0,sizeof a);
memset(vis,0,sizeof vis);
queue<int>q;
q.push(n);
while(!q.empty())
{
t=q.front();
q.pop();
if(t+1<N && vis[t+1]==0)
{
vis[t+1]=1;
q.push(t+1);
a[t+1]=a[t]+1;
}
if(t+1==k) {break;}
if(t-1>=0 && vis[t-1]==0)
{
vis[t-1]=1;
q.push(t-1);
a[t-1]=a[t]+1;
}
if(t-1==k) {break;}
if(t*2<N && vis[t*2]==0)
{
vis[t*2]=1;
q.push(t*2);
a[t*2]=a[t]+1;
}
if(t*2==k) {break;}
}
printf("%d\n",a[k]);
}
return 0;
}
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原文地址：http://blog.csdn.net/wust_zjx/article/details/45749045
