USACO section 2.1 Hamming Codes(直接枚举暴搜)

Hamming Codes
Rob Kolstad

Given N, B, and D: Find a set of N codewords (1 <= N <= 64), each of length B bits (1 <= B <= 8), such that each of the codewords is at least Hamming distance of D (1 <= D <= 7) away from each of the other codewords. The Hamming distance between a pair of codewords is the number of binary bits that differ in their binary notation. Consider the two codewords 0x554 and 0x234 and their differences (0x554 means the hexadecimal number with hex digits 5, 5, and 4):

        0x554 = 0101 0101 0100        0x234 = 0010 0011 0100Bit differences: xxx  xx

Since five bits were different, the Hamming distance is 5.

PROGRAM NAME: hamming

INPUT FORMAT

N, B, D on a single line

SAMPLE INPUT (file hamming.in)

16 7 3

OUTPUT FORMAT

N codewords, sorted, in decimal, ten per line. In the case of multiple solutions, your program should output the solution which, if interpreted as a base 2^B integer, would have the least value.

SAMPLE OUTPUT (file hamming.out)

0 7 25 30 42 45 51 52 75 7682 85 97 102 120 127

刚拿到题还不知道怎办，有想过暴搜，但怕暴，百度了下，发现真可以暴搜，就直接搜了明显是数据太弱了，

要是输入 126 30 6 就很久没反应了。

 

/*ID:nealgav1PROG:hammingLANG:C++*/#include<fstream>using namespace std;const int mm=1<<8;const int oo=1e9;int a,b,c,len;int hamm[79],lham;int hamming_length(int x,int y){  int z=x^y;int ans=0;  while(z>0)  {    if(z&1)    ans++;    z>>=1;  }  return ans;}void dfs(int dep){  if(dep==a)return;  int i,j;  for(i=1;i<len;i++)  {  bool flag=0;    for(j=0;j<lham;j++)    if(hamming_length(i,hamm[j])<c){flag=1;break;}    if(!flag){hamm[j]=i,lham++;break;}  }  dfs(dep+1);}int main(){  ifstream cin("hamming.in");  ofstream cout("hamming.out");  cin>>a>>b>>c;  len=1<<b;lham=1;hamm[0]=0;  dfs(1);  cout<<"0";  for(int i=1;i<lham;i++)  if(i%10==0)cout<<"\n"<<hamm[i];  else  cout<<" "<<hamm[i];  cout<<"\n";  return 0;}

 

 

 

Hamming Codes
Brian Jacokes

There are only a few tools we need to solve this problem. First of all, we can use basic techniques to find the Nth bit of a number M: counting the least significant bit as bit 0, the next bit as bit 1, etc., we can find the value of that bit through the expression

          int Nth_bit = (1 << N) & M;

In other words, we are shifting the number 1 left by N spaces, and then performing a binary AND on the resulting number with our original number, which will leave either a 1 in the Nth bit or a 0. So the first thing we have to do is find out the distance between each pair of numbers within the set of all numbers with B bits (0..2^B-1).

We also know that 0 can be one of the numbers in the set, because if the minimum number in the set were N instead of 0, we could XOR N to each number in the set and they would still be the same distance apart. The limits on the problem are low enough that we can do a DFS, and as soon as we hit a solution we can output it and exit.

#include <stdio.h>#include <stdlib.h>#include <iostream.h>#define MAX (1 << 8 + 1)#define NMAX 65#define BMAX 10#define DMAX 10int nums[NMAX], dist[MAX][MAX];int N, B, D, maxval;FILE *in, *out;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);        }    }}int main() {    in = fopen("hamming.in", "r");    out = fopen("hamming.out", "w");    int a, b, c;    fscanf(in, "%d%d%d", &N, &B, &D);    maxval = (1 << B);    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]++;        }    nums[0] = 0;    findgroups(1, 1);    return 0;}

 

 

 

USER: Neal Gavin Gavin [nealgav1]TASK: hammingLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 3336 KB]   Test 2: TEST OK [0.011 secs, 3336 KB]   Test 3: TEST OK [0.000 secs, 3336 KB]   Test 4: TEST OK [0.000 secs, 3336 KB]   Test 5: TEST OK [0.000 secs, 3336 KB]   Test 6: TEST OK [0.000 secs, 3336 KB]   Test 7: TEST OK [0.000 secs, 3336 KB]   Test 8: TEST OK [0.000 secs, 3336 KB]   Test 9: TEST OK [0.000 secs, 3336 KB]   Test 10: TEST OK [0.000 secs, 3336 KB]   Test 11: TEST OK [0.000 secs, 3336 KB]All tests OK.
Your program ('hamming') produced all correct answers!  This is yoursubmission #2 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 ----16 7 3------- test 2 ----2 7 7------- test 3 ----3 6 4------- test 4 ----5 5 1------- test 5 ----10 8 4------- test 6 ----20 6 2------- test 7 ----32 5 1------- test 8 ----50 8 2------- test 9 ----60 7 2------- test 10 ----16 8 3------- test 11 ----15 8 4

Keep up the good work!

Thanks for your submission!