Catch That Cow BFS

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本文探讨了农民约翰如何使用步行和瞬移两种方式，在最短时间内追捕到在数轴上某个位置的逃逸奶牛。通过最优路径规划，农民约翰成功在四分钟内追捕到了奶牛。

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Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 42562		Accepted: 13224

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK

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

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.

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
 struct node{
    int cur;
    int ts;
    node(int cur,int ts):cur(cur),ts(ts){}
 };
bool vis[300010];
int st,ed;

int bfs()
{
   if(st>ed) return st-ed;
   int len=ed-st;
    queue<node>que;
    que.push(node(st,0));
    vis[st]=1;
    while(que.size())
    {
        node tp=que.front(); que.pop();
        if(tp.cur==ed) return tp.ts;
        else if(tp.ts>len) continue;
        else{
                 if(tp.cur<ed&&vis[tp.cur+1]==0)
                 {
                      que.push(node(tp.cur+1,tp.ts+1));
                      vis[tp.cur+1]=1;
                 }
                  if(tp.cur-1>=0&&vis[tp.cur-1]==0)
                  {
                      que.push(node(tp.cur-1,tp.ts+1));
                      vis[tp.cur-1]=1;
                  }

                 if(tp.cur*2<=ed*2&&vis[tp.cur*2]==0)
                 {
                      que.push(node(tp.cur*2,tp.ts+1));
                      vis[tp.cur*2]=1;
                 }

          }
    }
}

int main()
{

     // freopen("in.in","r",stdin);
      while(~scanf("%d%d",&st,&ed))
      {
         memset(vis,0,sizeof(vis));
         cout<<bfs()<<endl;
      }
     return 0;
}