POJ 3278 Catch That Cow (BFS)

Catch That Cow

 

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 82733 Accepted: 25987

 

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 - 1 or + 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 100005

using 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;
}