Uva1368-DNA consensus string

Given m dna sequences with N lengths, find a DNA sequence so that the total Hamming distance to all sequences is as small as possible. The Hamming distance between two equal-length strings is equal to the number of different characters. Returns the minimum Lexicographic Order.

Idea: We can enumerate the letters at each position in sequence. To minimize the total Hamming, we need to take the same number of letters at each position. If the number is the same, we need to take the smallest Lexicographic Order.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 1005;const int N = 55;char str[N][MAXN], s[MAXN];int m, n, Min;int num[4];int Max(int a, int c, int g, int t) {    return max(a, max(c, max(g, t)));}int main() {    int cas;    scanf("%d", &cas);    while (cas--) {        scanf("%d%d", &m, &n);         for (int i = 0; i < m; i++)            scanf("%s", str[i]);        Min = 0;        int a, c, g, t;        for (int i = 0; i < n; i++) {             a = c = g = t = 0;            for (int j = 0; j < m; j++) {                if (str[j][i] == 'A')                    a++;                else if (str[j][i] == 'C')                    c++;                else if (str[j][i] == 'G')                    g++;                else if (str[j][i] == 'T')                    t++;             }            int k = Max(a, c, g, t);            Min += m - k;            if (k == a)                 s[i] = 'A';            else if (k == c)                 s[i] = 'C';            else if (k == g)                 s[i] = 'G';            else if (k == t)                 s[i] = 'T';         }        s[n] = '\0';        printf("%s\n",s);        printf("%d\n", Min);    }    return 0;}