第十四届华中科技大学程序设计竞赛 J-Various Tree 【BFS】【位运算】

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.
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. 大写的门

#include<bits/stdc++.h>
using namespace std;
const int MAX = 1e6 + 7;
struct node
{
    int x, s;
} op, uv;
int vis[MAX];
int f(int x)
{
    int ans = 0;
    int xxx = 1;
    while(xxx <= x)
    {
        if(x & xxx)
            ans++;
        xxx <<= 1;
    }
    return ans;
}
bool judge(int x)
{
    if(x >= 0 && x <= 1e7)
        return 1;
    return 0;
}
void bfs(int s, int t)
{
    queue <node> q;
    memset(vis, 0, sizeof vis);
    op.x = s;
    op.s = 0;
    vis[s] = 1;
    q.push(op);
    while(!q.empty())
    {
        op = q.front();
        if(op.x == t)
        {
            printf("%d\n", op.s);
            return;
        }
        q.pop();
        for(int i = 0; i < 4; i++)
        {
            if(i == 0)
            {
                uv.x = op.x + 1;
                if(!vis[uv.x] && judge(uv.x))
                {
                    vis[uv.x] = 1;
                    uv.s = op.s + 1;
                    q.push(uv);
                }
            }
            if(i == 1)
            {
                uv.x = op.x - 1;
                if(!vis[uv.x] && judge(uv.x))
                {
                    vis[uv.x] = 1;
                    uv.s = op.s + 1;
                    q.push(uv);
                }
            }
            if(i == 2)
            {
                int x = f(op.x);
                uv.x = op.x + x;
                if(!vis[uv.x] && judge(uv.x))
                {
                    vis[uv.x] = 1;
                    uv.s = op.s + 1;
                    q.push(uv);
                }
            }
            if(i == 3)
            {
                int x = f(op.x);
                uv.x = op.x - x;
                if(!vis[uv.x] && judge(uv.x))
                {
                    vis[uv.x] = 1;
                    uv.s = op.s + 1;
                    q.push(uv);
                }
            }
        }
    }
}
int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    bfs(n, m);
    return 0;
}