USACO Section 2.1 Hamming Codes

题目原文

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

分析

题目非常简单，直接从小到大枚举就可以了。计算两个整数的海明距离的时候，可以使用移位操作符来实现，具体代码如下：

int getHammingDistance(int a,int b){int dis=0;for (int i=0;i!=B;i++){if((1<<i & a) != (1<<i & b))dis++;}return dis;}

提交代码

/*ID: PROG: hammingLANG: C++*/#include <iostream>#include <fstream>#include <vector>#include <algorithm>#include <string>#include <math.h>#include <limits>#include <map>using namespace std;int N,B,D;int getHammingDistance(int a,int b){int dis=0;for (int i=0;i!=B;i++){if((1<<i & a) != (1<<i & b))dis++;}return dis;}int main(){ifstream cin("hamming.in");ofstream cout("hamming.out");cin >> N >> B >> D;vector<int> codes;codes.push_back(0);int temp=1;int count=1;while(count<N && temp<pow(2.0,B)){bool flag = true;for (int i=0;i!=codes.size();i++){flag = flag && (getHammingDistance(temp,codes[i])>=D);if(!flag)break;}if(flag){codes.push_back(temp);count++;}temp++;}for (int i=0;i!=codes.size()-1;i++){cout << codes[i];if(i%10==9)cout <<endl;elsecout << " ";}cout << codes[codes.size()-1] << endl;return 0;}

提交结果

TASK: hammingLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.011 secs, 3500 KB]   Test 2: TEST OK [0.008 secs, 3500 KB]   Test 3: TEST OK [0.005 secs, 3500 KB]   Test 4: TEST OK [0.008 secs, 3500 KB]   Test 5: TEST OK [0.005 secs, 3500 KB]   Test 6: TEST OK [0.005 secs, 3500 KB]   Test 7: TEST OK [0.008 secs, 3500 KB]   Test 8: TEST OK [0.008 secs, 3500 KB]   Test 9: TEST OK [0.011 secs, 3500 KB]   Test 10: TEST OK [0.005 secs, 3500 KB]   Test 11: TEST OK [0.003 secs, 3500 KB]All tests OK.

官方参考答案

#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;}

THE END