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

题意

农夫需要找到他的牛，农夫现在所在位置是N，牛在K点，假设牛是不动的，农夫有三种走法：

1. 向前走一步（N+1）
2. 向后走一步（N-1）
3. 跳到当前位置两倍处（2*N）

问，最少走多少次才能找到他的牛。

思路

简单的bfs，把农夫当前位置加入队列，然后出队判断每一个点，并模拟下一次的走法，注意的是，需要标记走过的点，否则会爆栈。

AC代码

#include <iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<queue>#include<math.h>#define MAXX 100005using namespace std;bool isvisted[MAXX];struct node{    int n;    int time;    node(int n,int time)    {        this->n=n;        this->time=time;    }};void bfs(int n,int k){    memset(isvisted,false,sizeof(isvisted));    node ans=node(n,0);    isvisted[n]=true;    queue<node>sk;    sk.push(ans);    while(!sk.empty())    {        node p=sk.front();        sk.pop();        isvisted[p.n]=true;     //标记走过的点        if(p.n==k)  //找到        {            ans=p;            break;        }        if(p.n+1<MAXX&&!isvisted[p.n+1])    //三种走法            sk.push(node(p.n+1,p.time+1));        if(p.n-1>=0&&!isvisted[p.n-1])            sk.push(node(p.n-1,p.time+1));        if(p.n*2<MAXX&&!isvisted[p.n*2])            sk.push(node(p.n*2,p.time+1));    }    printf("%d\n",ans.time);}int main(){    int n,k;    while(~scanf("%d%d",&n,&k))        bfs(n,k);    return 0;}