Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KB

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Problem 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.

Example Input

5 17

Example Output

4

题意大概就是：约翰丢了一头牛，这头牛和约翰在同一个数轴上，数轴大于0。约翰有两种交通方式，要么步行，一次最多只能前进一步或者退后一步；要么传送，只能传送为当前位置的2倍。给你约翰和牛的位置，问你约翰最少通过几次能找到牛？牛沉浸在自由的喜悦中浑然不知约翰在找她，随所以牛静止不动。

显然BFS，层数最少使得n==k

#include<cstdio>  #include<iostream>  #include<cstring>  #include<queue>  #include<algorithm>  using namespace std;    const int N = 200100;  int n, k;  struct node  {      int x, step;  };  queue<node> Q;  int vis[N];    void BFS()  {   //搜索关于X的三种状态      int X, STEP;      while(!Q.empty())  //队列非空     {          node t = Q.front(); //队头元素t出队         Q.pop();          X = t.x;          STEP = t.step;          if(X == k)  //如果当前位置和目的位置相等 直接输出step        {              printf("%d\n",STEP);              return ;          }          if(X >= 1 && vis[X - 1]==0) //要保证减1后有意义,所以要X >= 1，并且X-1未被访问         {              node temp;              vis[X - 1] = 1; //访问置1             temp.x = X - 1;  //当前位置             temp.step = STEP + 1;  //步数加1             Q.push(temp);  //进队         }          if(X <= k && vis[X + 1]==0)  //要保证减1后有意义当前位置比目的地小并未被访问         {              node temp;              vis[X + 1] = 1;              temp.x = X + 1;              temp.step = STEP + 1;              Q.push(temp);  //进队         }          if(X <= k && vis[X * 2]==0)  //要保证*2后有意义且未被访问         {              node temp;              vis[X * 2] = 1;              temp.x = 2 * X;              temp.step = STEP + 1;              Q.push(temp); //入队         }      }  }    int main()  {      while(~scanf("%d%d",&n,&k))      {          while(!Q.empty()) //依次弹出 Q.pop();          memset(vis,0,sizeof(vis));          vis[n] = 1;  //         node t;          t.x = n, t.step = 0;  //n是当前位置,step是走的步数         Q.push(t);  //t进队         BFS();      }      return 0;  }