POJ 3278 Catch That Cow（BFS）

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解决POJ 3278问题，利用BFS算法找到农夫约翰最快抓到不动的奶牛所需的时间。考虑不同情况下的最优策略。

题目链接：http://poj.org/problem?id=3278

题目大意:给定n和k，农夫在n位置，奶牛在k位置，奶牛不动，农夫只有三种走法，+1，-1，*2，问最短时间抓到奶牛

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

题目思路：最短路径，BFS，当k<=n时，只能一直减，答案是n-k，否则用BFS求最短，记得标记已经访问过的点

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 2e5+5;

int n,k;
int vis[N];

struct node
{
    int x,step;
}a,next;

bool check(int x)
{
    if(x<0||x>N||vis[x]==1)
        return false;
    return true;
}

int bfs()
{
    queue<node>q;
    a.x=n;
    a.step=0;
    vis[a.x]=1;
    q.push(a);
    while(!q.empty())
    {
        a=q.front();
        q.pop();
        if(a.x==k)
            return a.step;
        next.x=a.x*2;
        next.step=a.step+1;
        if(check(next.x))
        {
            vis[next.x]=1;
            q.push(next);
        }
        next.x=a.x+1;
        next.step=a.step+1;
        if(check(next.x))
        {
            vis[next.x]=1;
            q.push(next);
        }
        next.x=a.x-1;
        next.step=a.step+1;
        if(check(next.x))
        {
            vis[next.x]=1;
            q.push(next);
        }
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);

    while(~scanf("%d%d",&n,&k))
    {
        if(k<=n)
            printf("%d\n",n-k);
        else
        {
            MEM(vis,0);
            printf("%d\n",bfs());
        }
    }
    return 0;
}