POJ 3278 Catch That Cow(BFS基础)

题目大意：

　　给你两个数字，n和k，对于n有三种变化规则。

　　1.n->n-1;

　　2.n->n+1;

　　3.n->2*n;

　　使得n==k，问如何需要最少的操作次数。

解题思路：

　　凡是有关最少操作次数的问题，就归结为bfs来求解，这道题是一个三入口式的bfs。。。

在搜索的过程中，一定要注意是否有需要剪纸的必要，不然蛮力上的话，必然会导致RE。

代码：

# include<cstdio>
# include<iostream>
# include<algorithm>
# include<functional>
# include<cstring>
# include<string>
# include<cstdlib>
# include<iomanip>
# include<numeric>
# include<cctype>
# include<cmath>
# include<ctime>
# include<queue>
# include<stack>
# include<list>
# include<set>
# include<map>

using namespace std;

const double PI=4.0*atan(1.0);

typedef long long LL;
typedef unsigned long long ULL;

# define inf 999999999
# define MAX 100000+4

int step[MAX];
int book[MAX];
int n,k;
int ans;
queue<int>Q;

int bfs( int n,int k )
{
    int next,head;
    step[n] = 0;
    book[n] = 1;
    Q.push(n);
    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 = 2*head;
            if ( next>MAX||next<0 )
                continue;

            if ( book[next]==0 )
            {
                book[next] = 1;
                Q.push(next);
                step[next] = step[head]+1;
            }
            if ( next==k )
                return step[next];
        }

    }
}

int main(void)
{
    while ( cin>>n>>k )
    {
        while ( !Q.empty() )
        Q.pop();
        ans = 0;
        memset(step,0,sizeof(step));
        memset(book,0,sizeof(book));
        if ( n>=k )
            ans = n-k;
        else
            ans = bfs(n,k);
        cout<<ans<<endl;
    }


    return 0;
}