1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 829  Solved: 393
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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?

    农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．
    他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有
两种办法移动，步行和瞬移：步行每秒种可以让约翰从z处走到x+l或x-l处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．
    那么，约翰需要多少时间抓住那只牛呢？

Input

* Line 1: Two space-separated integers: N and K

    仅有两个整数N和K.

 

Output

* Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

    最短的时间．

Sample Input

5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.

Sample Output

4

OUTPUT DETAILS:

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.

HINT

 

Source

Silver

就是简单的bfs吧。然后就是开始不知道要最大到哪里就停止，然后想到其实很明显，每一个点到达另一个点都可以有两种方法，然后ac了

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#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int n,m,maxn;
const int nmax=100005;
int q[nmax],v[nmax],d[nmax];
int main(){
 scanf("%d%d",&n,&m);
 maxn=max(n,m)+1;
 memset(d,0x7f,sizeof(d));
 memset(v,0,sizeof(v));
 int l=0;int r=1;
    q[0]=n;
    v[n]=1;
    d[n]=0;
 while(l!=r){
  int tmp=q[l];
  l++;
  if(l==nmax) l=0;
  int dis=d[tmp]+1;
  if(tmp-1>=0&&d[tmp-1]>dis){
   d[tmp-1]=dis;
   if(tmp-1==m)
     break;
   if(!v[tmp-1]){
    v[tmp-1]=1;
    q[r++]=tmp-1;
    if(r==100000)
      r=0;
   }
  }
     if(tmp+1<=maxn&&d[tmp+1]>dis){
      d[tmp+1]=dis;
      if(tmp+1==m)
        break;
      if(!v[tmp+1]){
       v[tmp+1]=1;
       q[r++]=tmp+1;
       if(r==100000)
         r=0;
      }
     }
     if(tmp*2<=maxn&&d[tmp*2]>dis){
      d[tmp*2]=dis;
      if(tmp*2==m)
        break;
      if(!v[tmp*2]){
       v[tmp*2]=1;
       q[r++]=tmp*2;
       if(r==100000)
         r=0;
      }
     }
 }
 printf("%d\n",d[m]);
 return 0;
}

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