M - DNA Consensus String

The Hamming distance is the number of different characters at each position from two strings of equal length. For example, assume we are given the two strings “AGCAT” and “GGAAT.” The Hamming distance of these two strings is 2 because the 1st and the 3rd characters of the two strings are different. Using the Hamming distance, we can define a representative string for a set of multiple strings of equal length. Given a set of strings S = {s1,…,sm} of length n, the consensus error between a string y of length n and the set S is the sum of the Hamming distances between y and each si in S. If the consensus error between y and S is the minimum among all possible strings y of length n, y is called a consensus string of S. For example, given the three strings “AGCAT” “AGACT” and “GGAAT” the consensus string of the given strings is “AGAAT” because the sum of the Hamming distances between “AGAAT” and the three strings is 3 which is minimal. (In this case, the consensus string is unique, but in general, there can be more than one consensus string.) We use the consensus string as a representative of the DNA sequence. For the example of Figure 2 above, a consensus string of gene X is “GCAAATGGCTGTGCA” and the consensus error is 7.

Input
Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with a line containing two integers m and n which are separated by a single space. The integer m (4 ≤ m ≤ 50) represents the number of DNA sequences and n (4 ≤ n ≤ 1000) represents the length of the DNA sequences, respectively. In each of the next m lines, each DNA sequence is given.
Output
Your program is to write to standard output. Print the consensus string in the first line of each case and the consensus error in the second line of each case. If there exists more than one consensus string, print the lexicographically smallest consensus string.
Sample Input
3
5 8
TATGATAC
TAAGCTAC
AAAGATCC
TGAGATAC
TAAGATGT
4 10
ACGTACGTAC
CCGTACGTAG
GCGTACGTAT
TCGTACGTAA
6 10
ATGTTACCAT
AAGTTACGAT
AACAAAGCAA
AAGTTACCTT
AAGTTACCAA
TACTTACCAA
Sample Output
TAAGATAC
7
ACGTACGTAA
6
AAGTTACCAA
12

#include <stdio.h>
#include <string.h>
char p[51][1010];
int g[90];
char bi[30];
int k=0;
char search1(int n)
{
    int max1=bi[0];
    for(int a=0;a<n;a++)
    {
        if(g[bi[a]]>g[max1])max1=bi[a];
        else if(g[max1]==g[bi[a]])
        {
            if(max1>bi[a])max1=bi[a];
        }
    }
    k=g[max1];
    return max1;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int top=0;
        int i=0;
        memset(p, 0, sizeof(p));
        memset(g, 0, sizeof(g));
        int n, m;
        scanf("%d%d", &n, &m);
        for(int a=0;a<n;a++)
        {
            scanf("%s", p[a]);
        }
        for(int b=0;b<m;b++)
        {
            i=0;
            memset(g, 0, sizeof(g));
            for(int a=0;a<n;a++)
            {
                if(!g[p[a][b]])
                {
                    g[p[a][b]]++;
                    bi[i++]=p[a][b];
                }
                else g[p[a][b]]++;
            }
             printf("%c", search1(i));
             top+=(n-k);
        }
        printf("\n");
        printf("%d\n", top);
    }
    return 0;
}