POJ3278-Catch That Cow(bfs)

POJ3278 原题链接：http://poj.org/problem?id=3278

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.

题目大意:

John要找牛,输入John和牛的位置。John有三种走法:

1.前进一步耗时1min(5->6)

2.后退一步耗时1min(5->4)

3.前进到两倍于现在的位置(5->10)

求耗时最短所需时间.

思路:bfs(队列)

代码如下:

#include<iostream>#include<queue>#include<cstring>#include<cstdio>using namespace std;const int maxn=100001;bool vis[maxn];int step[maxn];queue <int> q;int bfs(int n,int k){    int head,next;    q.push(n);    step[n]=0;    vis[n]=true;    while(!q.empty())    {        head=q.front();        q.pop();        for(int i=0;i<3;i++)        {            if(i==0) next=head-1;            else if(i==1) next=head+1;            else next=head*2;            if(next<0 || next>=maxn) continue;            if(!vis[next])            {                q.push(next);                step[next]=step[head]+1;                vis[next]=true;            }            if(next==k) return step[next];        }    }}int main(){    int n,k;    while(cin>>n>>k)    {        memset(step,0,sizeof(step));        memset(vis,false,sizeof(vis));        while(!q.empty()) q.pop();        if(n>=k) printf("%d\n",n-k);        else printf("%d\n",bfs(n,k));    }    return 0;}

参考文章:http://blog.csdn.net/freezhanacmore/article/details/8168265

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