| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 32071 | Accepted: 9866 |
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
代码：
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<queue>
using namespace std;
int n,k;
queue<int>Q;
#define N 100005
int visit[N];
int num[N];
int BFS(int x)
{
Q.push(x);
visit[x]=1;
int a,b;
while(!Q.empty())
{
a=Q.front();
Q.pop();
for(int i=0;i<3;i++)
{
if(i==0)
b=a+1;
else if(i==1)
b=a-1;
else
b=2*a;
if(b<0||b>N)
continue;
if(!visit[b])
{
Q.push(b);
visit[b]=1;
num[b]=num[a]+1;
if(b==k)
return num[b];
}
}
}
}
int main()
{
scanf("%d%d",&n,&k);
memset(num,0,sizeof(num));
memset(visit,0,sizeof(visit));
if(n>=k)
printf("%d\n",n-k);
else
printf("%d\n",BFS(n));
//system("pause");
return 0;
}
