POJ_3278 Catch That Cow(BFS)

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.
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.

题目在这里～
题目描述：
给出一个数字，可执行三种操作，+1，-1，×2.问最少经过多少步能到达目标点。
题解：
因为思路很简单，就是一道BFS，手写DFS错了很久，关键在于不好记忆化。放到队列里面标记很好的解决了这个问题。已经在队列里面的点不需要继续放入，防止了重复计算。
代码：

#include <iostream>#include <math.h>#include <cstring>#include <algorithm>#include <queue>using namespace std;struct node{    int val;    int cos;};int num;int N,K;queue<node> q;int show[100000+2];void DFS(int x,int time);int main(){    while( cin>>N>>K )    {        memset(show,0,sizeof(show));        show[N]=1;        num = -1;        node a,tem;        show[N]=1;        a.val=N;        a.cos=0;        q.push(a);        while( !q.empty() )        {            node t;            tem = q.front();            q.pop();            if( tem.val == K )            {                num=tem.cos;                break;            }            if( tem.val + 1 <= 100000 && show[tem.val+1]!=1 )            {                t.val=tem.val+1;                t.cos=tem.cos+1;                q.push(t);                show[tem.val+1]=1;            }            if( tem.val-1 >= 0 && show[tem.val-1]!=1)            {                t.val=tem.val-1;                t.cos=tem.cos+1;                q.push(t);                show[tem.val-1]=1;            }            if( tem.val*2 <= 100000 && show[tem.val*2]!=1 )            {                t.val=tem.val*2;                t.cos=tem.cos+1;                q.push(t);                show[tem.val*2]=1;            }        }        cout<<num<<endl;    }    return 0;}