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POJ 3278 Catch That Cow

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Catch That Cow Time Limit:2000MS Memory Limit:65536K Total 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 ≤


Catch That Cow


Time Limit: 2000MS

 

Memory Limit: 65536K

Total 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 - 1 or + 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;
}




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