第十四届华中科技大学程序设计竞赛--J Various Tree

链接：https://www.nowcoder.com/acm/contest/106/J
来源：牛客网

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

It’s universally acknowledged that there’re innumerable trees in the campus of HUST.

And there are many different types of trees in HUST, each of which has a number represent its type. The doctors of biology in HUST find 4 different ways to change the tree’s type x into a new type y:

1.    y=x+1

2.    y=x-1

3.    y=x+f(x)

4.    y=x-f(x)

The function f(x) is defined as the number of 1 in x in binary representation. For example, f(1)=1, f(2)=1, f(3)=2, f(10)=2.

Now the doctors are given a tree of the type A. The doctors want to change its type into B. Because each step will cost a huge amount of money, you need to help them figure out the minimum steps to change the type of the tree into B. 

Remember the type number should always be a natural number (0 included).

输入描述:

One line with two integers A and B

, the init type and the target type.

输出描述:

You need to print a integer representing the minimum steps.

输入例子:

5 12

输出例子:

3

-->

示例1

输入

5 12

输出

3

说明

The minimum steps they should take: 5->7->10->12. Thus the answer is 3.

开始以为广搜会t，然而并不会

/*
    data:2018.5.20
    author:gsw
    link:https://www.nowcoder.com/acm/contest/106#question
*/
#define ll long long
#define IO ios::sync_with_stdio(false);

#include<math.h>
#include<stdio.h>
#include<queue>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define maxn 1000005

int brainy[maxn];
int vis[maxn];
int getbrainy(int a)
{
    int ans=;
    while(a>)
    {
        if(a&)ans++;
        a=a>>;
    }
    return ans;
}
void init()
{
    for(int i=;i<maxn;i++)
        brainy[i]=getbrainy(i);
    memset(vis,,sizeof(vis));
}
class Step
{
    public:
        int x,st;
};
void bfs(int a,int b)
{
    Step be,ne;
    be.x=a;be.st=;
    queue<Step> q;
    q.push(be);
    vis[be.x]=;
    while(!q.empty())
    {
        be=q.front();
        q.pop();
        if(be.x==b)
        {
            cout<<be.st<<endl;
            return;
        }

        if(!vis[be.x+])
        {
            vis[be.x+]=;
            ne.x=be.x+;ne.st=be.st+;
            q.push(ne);
        }
        if((be.x-brainy[be.x])>=&&!vis[be.x-brainy[be.x]])
        {
            vis[be.x-brainy[be.x]]=;
            ne.x=be.x-brainy[be.x];ne.st=be.st+;
            q.push(ne);
        }
        if((be.x-)>=&&!vis[be.x-])
        {
            vis[be.x-]=;
            ne.x=be.x-;ne.st=be.st+;
            q.push(ne);
        }
        if(!vis[be.x+brainy[be.x]])
        {
            vis[be.x+brainy[be.x]]=;
            ne.x=be.x+brainy[be.x];ne.st=be.st+;
            q.push(ne);
        }
    }
}
int main()
{
    int a,b;
    init();
    scanf("%d%d",&a,&b);
    bfs(a,b);
}