第十四届华中科技大学-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.

解题思路：BFS，广度搜索，从a出发，找到b所需要的最短步数

AC代码;

#include<bits/stdc++.h>
using namespace std;
int a, b;
int ans = 0;
const int maxn = 1e6 + 4;
bool fl[maxn];
struct Node
{
	int x, step;
};
queue<Node>Q;
int f(int x)
{
	int ans = 0;
	while (x)
	{
		if (x & 1) ans++;
		x >>= 1;
	}
	return ans;
}
int fun(int i,int x)
{
	if (x == 0) return i - 1;
	if (x == 1) return i + 1;
	if (x == 2) return i + f(i);
	if (x == 3) return i - f(i);
	else return 0;
}
int main()
{
	int flag = 0;
	cin >> a >> b;
	memset(fl, 0, sizeof(fl));
	Node now;
	now.x = a;
	now.step = 0;
	Q.push(now);
	int ans = 0;
	if (now.x == b)
	{
		cout << now.step << endl;
		return 0;
	}
	while (!Q.empty())
	{
		Node q = Q.front();
		fl[q.x] = 1;
		Q.pop();
		for (int i = 0; i < 4; i++)
		{
			int temp = fun(q.x, i);
			if (temp == b)
			{
				ans = q.step + 1;
				cout << ans << endl;
				return 0;
			}
			if (temp >= 0 && fl[temp] == 0)
			{
				Node next;
				next.x = temp;
				next.step = q.step + 1;
				fl[temp] = 1;
				Q.push(next);
			}
		}
	}
	cout << ans << endl;
	return 0;
}