USACO 2.1 Hamming Codes-蒲公英云

USACO 2.1 Hamming Codes

Problem 94: 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 0100
Bit 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
Type:DFS
思路：先计算出0-2^B-1,任意两个数的距离，可以动态的求解。再根据：0必然在最后的结果中，因为如果最小的数是b0，
让每个数减去b0，仍然符合距离的限制条件。所以只需从0开始按递增的顺序搜索，一旦搜到答案，就可exit;

代码

#include < stdio.h >
#include < stdlib.h >
#include < iostream.h >
#define MAX (1 << 8 + 1)
#define NMAX 65
#define BMAX 10
#define DMAX 10

int 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 ;
}

转载于//www.cnblogs.com/superbin/archive/2010/06/03/1750977.html