POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 44220 Accepted: 13814

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.

Source

USACO 2007 Open Silver

大致题意：

给定两个整数n和k

通过 n+1或n-1 或n*2 这3种操作，使得n==k

输出最少的操作次数

 

#include <stdio.h>#include <string.h>#include <stdlib.h>struct node{    int x,ans;}q[1000005];int jx[]={-1,1};int n,k;int v[1000005];void BFS(){     struct node t,f;     int i;     int e=0,s=0;     t.x=n;     v[t.x]=1;     t.ans=0;     q[e++]=t;     while(s<e)    {        t=q[s++];        if(t.x==k)        {             printf("%d\n",t.ans);             break;        }        for(i=0;i<3;i++)        {             if(i==2)             f.x=t.x*2;             else f.x=t.x+jx[i];             if(v[f.x]==0&&f.x>=0&&f.x<=100000)             {                 f.ans=t.ans+1;                 q[e++]=f;                 v[f.x]=1;             }        }    }}int main(){    while(scanf("%d%d",&n,&k)!=EOF)    {        memset(v,0,sizeof(v));        BFS();    }    return 0;}