Catch That Cow(广度优先搜索_bfs)



Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 48036 Accepted: 15057

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

4

题意：输入两个数n，k。求从n到k最少走多少步。可以前进1后退1或者当前的位置*2；

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>using namespace std;struct node{    int x;//当前位置    int ans;//走的步数}q[1000010];int vis[1000010];//标记变量，该点是否被访问；int jx[]={-1,1};//后退1或者前进1；struct node t,f;int n,k;void bfs(){     int i;     int s=0,e=0;//指针模拟队列。往队列加e++ 往队列里提出数s++     memset(vis,0,sizeof(vis));     t.x=n;//当前初始位置     vis[t.x]=1;//标记为1代表访问过；     t.ans=0;//初始位置步数为0；     q[e++]=t;//把当前步数加人队列     while(s<e)//当队列不为空    {        t=q[s++];//提出        if(t.x==k)//如果该数正好等于目标位置直接输出步数        {             printf("%d\n",t.ans);             break;        }        for(i=0;i<3;i++)//i=0后退一步，i=1前进一步，i=2此时的位置*2；        {             if(i==2)             {                 f.x=t.x*2;             }             else             {                 f.x=t.x+jx[i];             }             if(f.x>=0&&f.x<=100000&&!vis[f.x])             {                 f.ans=t.ans+1;                 q[e++]=f;                 vis[f.x]=1;             }        }    }}int main(){    while(~scanf("%d %d",&n,&k))    {        bfs();    }    return 0;}