POJ 3278 Catch That Cow

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本文介绍了一个经典的算法问题——如何使用广度优先搜索(BFS)策略来找到从起点到终点的最短路径。通过一个有趣的场景设定，即农夫追赶逃跑的奶牛，详细解释了BFS的应用及实现过程。

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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.

题意：给你一个数n让你按照规则走到p，要求步数最少。
用bfs即可，注意边界，其他。。就是模板bfs。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
queue<int>q;
int n,k,stp[100001];
int bfs(int s,int e)
{
    if(s==e)
        return 0;
    memset(stp,-1,sizeof(stp));
    stp[s]=0;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front(),v;
        q.pop();
        for(int i=1;i<=3;i++)
        {
            if(i==1)
                v=u-1;
            if(i==2)
                v=u+1;
            if(i==3)
                v=u*2;
            if(v<0||v>100000||stp[v]>=0)
                continue;
            stp[v]=stp[u]+1;
            q.push(v);
            if(v==e)
                return stp[v];
        }
    }
}
int main()
{
    scanf("%d%d",&n,&k);
    printf("%d\n",bfs(n,k));
    return 0;
}