poj3278 Catch That Cow

POJ2378 Catch That Cow 简单BFS

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Catch That Cow
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

题目大意：给你两个数n和k,n表示农夫的位置，k表示奶牛的位置，现在农夫去抓牛，农夫可以选择x-1,x+1,x*2得位置变化去抓牛，问最快可以几步抓到。

分析：问的是最快的，基本上就是bfs无疑了，确定 bfs的退出条件就是位置转移到k 也就是 x经过变化后 达到 x==k了。下边就可以写BFS了，用一个数组step[i]表示由x变到i的最小步数，所以 最后答案应该是step[k]。需要注意的是在判断边界情况的时候，最大的不是2*k 我的很多队友都是写这写挂的，因为有可能n比k还有大，所以我设非常的大的数，来判断边界。

#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];
        }
    }
    return -1;
}
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;
}