POJ 3278-Catch That Cow（BFS-一维广搜）-蒲公英云

POJ 3278-Catch That Cow（BFS-一维广搜）

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 82618		Accepted: 25955

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

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

题目意思：

在一维坐标上，奶牛丢在了K位置，农夫从N位置去找牛。

农夫只能有三种走法：①后退一步；②前进一步；③瞬移到二倍N的位置。

每次使用一种走法花费一分钟，求找到牛要花多久。

解题思路：

BFS，分别对三种走法广搜，记录走到当前位置花费多长时间，直到N==K。

#include <iostream>
#include <cstring>
#include <cstdio>
#include<queue>
using namespace std;
const int maxn=3000000;
int n,k;
int ans=0;
bool vis[maxn];//记录是否访问
int pace[maxn];//记录走过的所有位置
int head,pos;//当前位置和要走到的位置
queue <int> q;
int bfs()
{
    q.push(n);
    while(!q.empty())
    {
        head=q.front();
        q.pop();
        for(int i=0; i<3; ++i)//有三种走法
        {
            if(i==0) pos=head-1;//后退一步
            else if(i==1) pos=head+1;//前进一步
            else pos=head*2;//二倍步数
            if(pos<0||pos>100000) continue;//超过范围
            if(vis[pos]==false)//未访问过就加入队列
            {
                q.push(pos);
                pace[pos]=pace[head]+1;//记录到达当前位置走过多少步
                //cout<<pos<<endl;
                vis[pos]=true;
            }
            if(pos==k) return pace[pos];//找到奶牛
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while(cin>>n>>k)//农夫、奶牛位置
    {
        memset(vis,false,sizeof(vis));
        if(n>=k) cout<<n-k<<endl;//如果比目标点远，就只能一步步后退挪回来
        else cout<<bfs()<<endl;
    }
    return 0;
}
/*
5 17
*/