poj3278 catch that cow
Catch That Cow
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 43653 | Accepted: 13605 |
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
#include<stdio.h> #include<string.h> int front,rear,visit[100001],bu[100001],m,n; typedef struct { int w; }sqtype; sqtype sq[100001]; void bfs() { int x; front=rear=0; sq[0].w=m; bu[m]=0; visit[m]=1; while(front<=rear) { x=sq[front].w; if(visit[n]!=0) break; if(x-1>=0&&visit[x-1]==0) { rear++; sq[rear].w=x-1; bu[x-1]=bu[x]+1; visit[x-1]=1; } if(x+1<=100001&&visit[x+1]==0) { rear++; sq[rear].w=x+1; bu[x+1]=bu[x]+1; visit[x+1]=1; } if(x*2<=100001&&visit[x*2]==0) { rear++; sq[rear].w=x*2; bu[x*2]=bu[x]+1; visit[x*2]=1; } front++; } } main() { scanf("%d%d",&m,&n); memset(visit,0,sizeof(visit)); bfs(); printf("%d",bu[n]); }