Various Tree

好久没做题了...华科的题  考bfs

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

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

5 12

输出

3

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=2e6+5;
int vis[maxn];
struct node{
int weigh;
int step;
};
int f(int a)
{
int ans=0;
// cout<<a<<":";
while(a)
{
if(a&1)
ans++;

a>>=1;
}

// cout<<ans<<endl;
return ans;
}


int bfs(int a,int b)
{
node p;
p.weigh=a;
p.step=0;
vis[p.weigh]=1;
queue<node>s;
s.push(p);

node tmp;


while(!s.empty())
{

p=s.front();
if(p.weigh==b)
return p.step;

s.pop();

int x[4]={-1,1,f(p.weigh),-f(p.weigh)};

for(int i=0;i<4;i++)
{
tmp.weigh=p.weigh+x[i];
if(tmp.weigh<=maxn&&tmp.weigh>=0&&vis[tmp.weigh]==0)
{


vis[tmp.weigh]=1;
tmp.step=p.step+1;
s.push(tmp);
}

}

}
   return -1;

}
int main()
{
int n,m;
while(cin>>n>>m)
{
// cout<<"**"<<endl;
memset(vis,0,sizeof(vis));
int sum=bfs(n,m);

cout<<sum<<endl;
}


return 0;
 }