[USACO][枚举]Hamming Code

题意：

给出N，B，D，要求输出N个十进制数字，他们之间的Hamming距离在长度为B位的时候都等于D。

思路：

感觉图论就是一个，允许我们记忆化的边权好工具！（废话用edges存边了还不是当然的）

代码：

/*
ID :ggy_7781
TASK :hamming
LANG :C++11
*/

#include <bits/stdc++.h>
#define maxB 8
#define maxN 64
#define maxD 7
using namespace std;

int N,B,D;
int maxn;
int getDist(int x,int y)
{
    int tmp = (x^y);
    int cnt = 0;
    while(tmp)
    {
        cnt += (tmp&1);
        tmp = tmp >> 1;
    }
    return cnt;
}
int one[256][256];
int main(void)
{
    freopen("hamming.in","r",stdin);
    freopen("hamming.out","w",stdout);
    cin>>N>>B>>D;
    maxn = pow(2,B) ;
    for(int i = 0;i < maxn;i ++)
    {
        for(int j = 0;j <maxn;j ++)
        {
            one[i][j] = getDist(i,j);
        }
    }
    int pos = 0;
    vector<int> ret;
    ret.push_back(pos);
    while(ret.size() < N)
    {
        for(int i =pos + 1 ;i <maxn;i ++)
        {
            bool ok = true;
            for(int j = 0;j <ret.size();j ++)
            {
                if(one[ret[j]][i] < D)
                {
                    ok = false;
                    break;
                }
            }
            if(ok)
            {
                ret.push_back(i);
                pos = i;
                break;
            }
        }
    }
    int cnt = 0;
    bool bgn = true;
    bool edd = false;
    for(int i =0 ;i <N;i ++){
        cnt ++;
        if(bgn)
        {
            bgn = false;
            edd = false;
        }else
            cout<<" ";
        cout<<ret[i];
        if(cnt == 10)
        {
            bgn = true;
            edd = true;
            cnt = 0;
            cout<<endl;
        }
    }
    if(!edd)
        cout<<endl;
    return 0;
}