USACO Hamming Codes DFS 构造

我还是用了很朴素的暴力匹配A了这题，不得不感叹USACO时间放的好宽...

/*
ID: wushuai2
PROG: hamming
LANG: C++
*/
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int M = 660000         ;
const ll P = 10000000097ll   ;
const int INF = 0x3f3f3f3f   ;
const int MAX_N = 20         ;
const int MAXSIZE = 101000000;

int n, b, d;
int ans[80];

bool func(int b){
int i, j, cnt;
int t1[50], t2[50];
memset(t2, 0, sizeof(t2));
while(b){
t2[++t2[0]] = b % 2;
b = (b - b % 2) / 2;
}
for(int k = 1; k <= ans[0]; ++k){
cnt = 0;
int a = ans[k];
memset(t1, 0, sizeof(t1));
while(a){
t1[++t1[0]] = a % 2;
a = (a - a % 2) / 2;
}
for(i = 1; i <= Max(t1[0], t2[0]); ++i){
if(t1[i] != t2[i])  ++cnt;
}
if(cnt < d)    return false;
}
return true;
}

int main() {
ofstream fout ("hamming.out");
ifstream fin ("hamming.in");
int i, j, k, t, n, s, c, w, q;
fin >> n >> b >> d;
++ans[0];
ans[1] = 0;
int num = 1;
while(ans[0] <= n){
if(func(num)){
++ans[0];
ans[ans[0]] = num;
}
++num;
}
for(i = 1; i < ans[0]; ++i){
fout << ans[i];
if(i == ans[0] - 1){
break;
}
if(i % 10 == 0) fout << endl;
else    fout << ' ';
}
fout << endl;

fin.close();
fout.close();
return 0;
}

View Code
不过看了官方题解觉得很不错，来分享一下

for (a = 0; a < maxval; a++)
for (b = 0; b < maxval; b++) {
dist[a][b] = 0;
for (c = 0; c < B; c++)
if (((1 << c) & a) != ((1 << c) & b))
dist[a][b]++;
}

通过以上这段代码可以找到所有1 << B 中所有数的关系，就是二进制下不同的位数

然后通过一个DFS 来找可行对

void findgroups(int cur, int start) {
int a, b, canuse;
char ch;
if (cur == N) {
for (a = 0; a < cur; a++) {
if (a % 10)
fprintf(out, " ");
fprintf(out, "%d", nums[a]);
if (a % 10 == 9 || a == cur-1)
fprintf(out, "\n");
}
exit(0);
}
for (a = start; a < maxval; a++) {
canuse = 1;
for (b = 0; b < cur; b++)
if (dist[nums[b]][a] < D) {
canuse = 0;
break;
}
if (canuse) {
nums[cur] = a;
findgroups(cur+1, a+1);
}
}
}

不难得出，核心代码很短也很好写

找到全部N个数之后输出一下就可以了