POJ 3278 Catch That Cow(BFS,板子题)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 88732		Accepted: 27795

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.

Source

USACO 2007 Open Silver

题目链接：http://poj.org/problem?id=3278

题意：有一个农民和一头牛，他们在一个数轴上，牛在k位置保持不动，农户开始时在n位置。设农户当前在M位置，每次移动时有三种选择：1.移动到M-1；2.移动到M+1位置；3.移动到M*2的位置。问最少移动多少次可以移动到牛所在的位置。所以可以用广搜来搜索这三个状态，直到搜索到牛所在的位置。

分析：BFS的板子题，用队列做，第一次学队列，感觉自己好弱啊！现在才学，算法真的好难学，学了又怕不会用，现在不敢学太快了！

下面每一步代码我都给出了详细解释，一看就懂！

#include<iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include<queue>
#define MAX 100001
using namespace std;
queue<int> q;//使用前需定义一个queue变量，且定义时已经初始化
bool visit[MAX];//访问空间
int step[MAX];       //记录步数的数组不能少
bool bound(int num)//定义边界函数
{
    if(num<0||num>100000)
        return true;
    return false;
}
int BFS(int st,int end)
{
    queue<int> q;//使用前需定义一个queue变量，且定义时已经初始化
    int t,temp;
    q.push(st);//进队列
    visit[st]=true;
    while(!q.empty())//重复使用时，用这个初始化
    {
        t=q.front();//得到队首的值
        q.pop();//出队列,弹出队列的第一个元素，并不会返回元素的值
        for(int i=0;i<3;++i) //三个方向搜索
        {
            if(i==0)
                temp=t+1;
            else if(i==1)
                temp=t-1;
            else
                temp=t*2;
            if(bound(temp))         //越界
                continue;
            if(!visit[temp])//访问空间未被标记
            {
                step[temp]=step[t]+1;
                if(temp==end)
                    return step[temp];
                visit[temp]=true;//标记此点
                q.push(temp);//将temp元素接到队列的末端；
            }
        }
    }
}
int main()
{
    int st,end;
    while(scanf("%d%d",&st,&end)!=EOF)
    {
        memset(visit,false,sizeof(visit));//visit数组进行初始化操作
        if(st>=end)
            cout<<st-end<<endl;
        else
            cout<<BFS(st,end)<<endl;
    }
    return 0;
}