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
这题可以用宽搜，但要注意剪枝，否则队列会爆。
#include<stdio.h>
#include<string.h>
int n,m,x2,x1;
int q[11111111],b[200005];
void bfs()
{
int i,j,front=1,rear=1,x,xx;
memset(q,0,sizeof(q));memset(b,0,sizeof(b));
b[x1]=0;
q[front]=x1;
while(front<=rear){
x=q[front];
if(x==x2)break;
front++;
if(x+1<=100000 && b[x+1]==0){
rear++;q[rear]=x+1;b[x+1]=b[x]+1;
}
if(2*x<=200000 && b[2*x]==0){
rear++;q[rear]=2*x;b[2*x]=b[x]+1;
}
if(x>=1 && b[x-1]==0 ){
rear++;q[rear]=x-1;b[x-1]=b[x]+1;
}
//
}
return;
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
x1=n;x2=m;
bfs();
printf("%d\n",b[x2]);
}
return 0;
}