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=0;
while(a>0)
{
if(a&1)ans++;
a=a>>1;
}
return ans;
}
void init()
{
for(int i=1;i<maxn;i++)
brainy[i]=getbrainy(i);
memset(vis,0,sizeof(vis));
}
class Step
{
public:
int x,st;
};
void bfs(int a,int b)
{
Step be,ne;
be.x=a;be.st=0;
queue<Step> q;
q.push(be);
vis[be.x]=1;
while(!q.empty())
{
be=q.front();
q.pop();
if(be.x==b)
{
cout<<be.st<<endl;
return;
}
if(!vis[be.x+1])
{
vis[be.x+1]=1;
ne.x=be.x+1;ne.st=be.st+1;
q.push(ne);
}
if((be.x-brainy[be.x])>=0&&!vis[be.x-brainy[be.x]])
{
vis[be.x-brainy[be.x]]=1;
ne.x=be.x-brainy[be.x];ne.st=be.st+1;
q.push(ne);
}
if((be.x-1)>=0&&!vis[be.x-1])
{
vis[be.x-1]=1;
ne.x=be.x-1;ne.st=be.st+1;
q.push(ne);
}
if(!vis[be.x+brainy[be.x]])
{
vis[be.x+brainy[be.x]]=1;
ne.x=be.x+brainy[be.x];ne.st=be.st+1;
q.push(ne);
}
}
}
int main()
{
int a,b;
init();
scanf("%d%d",&a,&b);
bfs(a,b);
}
原文地址：https://www.cnblogs.com/fantastic123/p/9063441.html