Catch That Cow
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.
/*广度优先遍历*/
#include<stdio.h>
#include<string.h>
int k;
int vis[10000];
typedef struct node
{
int pos;
int step;
}node;
node cow[10005];
int bfs(int n)
{
node temp;
int head=0,rear=0;
vis[n]=1;
cow[rear].pos=n;
cow[rear++].step=0;
while(head<rear)
{
temp=cow[head++];
if( temp.pos>=0 && temp.pos+1<= 100000 && !vis[temp.pos+1]){
cow[rear].pos = temp.pos + 1;
cow[rear++].step = temp.step + 1;
vis[temp.step+1] = 1;
if( temp.pos + 1 == k) return temp.step + 1;
}
if( temp.pos>0 && temp.pos-1<=100000 && !vis[temp.pos-1]){
cow[rear].pos = temp.pos - 1;
cow[rear++].step = temp.step + 1;
vis[temp.step+1] = 1;
if( temp.pos - 1 == k) return temp.step + 1;
}
if( temp.pos>0 && temp.pos*2<= 100000 && !vis[temp.pos*2]){
cow[rear].pos = temp.pos *2;
cow[rear++].step = temp.step + 1;
vis[temp.step+1] = 1;
if( temp.pos *2 == k) return temp.step + 1;
}
}
}
int main()
{
int n;
while(scanf("%d%d",&n,&k)!=EOF)
{
memset(vis,0,sizeof(vis));
if(k<=n) printf("%d\n",n-k);
else printf("%d\n",bfs(n));
}
return 0;
}