DNA Consensus String UVA - 1368-蒲公英云

DNA Consensus String UVA - 1368

DNA (Deoxyribonucleic Acid) is the molecule which contains
the genetic instructions. It consists of four different
nucleotides, namely Adenine, Thymine, Guanine, and Cytosine
as shown in Figure 1. If we represent a nucleotide by
its initial character, a DNA strand can be regarded as a long
string (sequence of characters) consisting of the four characters
A, T, G, and C. For example, assume we are given some
part of a DNA strand which is composed of the following sequence
of nucleotides:
“Thymine-Adenine-Adenine-Cytosine-Thymine-Guanine-CytosineCytosine-Guanine-Adenine-Thymine”
Then we can represent the above DNA strand with the
string “TAACTGCCGAT.”
The biologist Prof. Ahn found that a gene X commonly
exists in the DNA strands of five different kinds of animals,
namely dogs, cats, horses, cows, and monkeys. He also discovered
that the DNA sequences of the gene X from each animal
were very alike. See Figure 2.
DNA sequence of gene X
Cat: GCATATGGCTGTGCA
Dog: GCAAATGGCTGTGCA
Horse: GCTAATGGGTGTCCA
Cow: GCAAATGGCTGTGCA
Monkey: GCAAATCGGTGAGCA
Figure 2. DNA sequences of gene X in five animals.
Prof. Ahn thought that humans might also have the gene X and decided to search for the DNA
sequence of X in human DNA. However, before searching, he should define a representative DNA
sequence of gene X because its sequences are not exactly the same in the DNA of the five animals. He
decided to use the Hamming distance to define the representative sequence.
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
题意：给你几个字符串，找到一个字符串，使其到所有字符串的总Hamming的距离尽量小，Hamming就是字符不同的位置个数。如有多解输出字典序最小的。和题目有关的就是序列只有 AGCT, 找每一列最多的那个。别怕麻烦。

#include <iostream>  
#include <iomanip>  
#include <string>  
#include <cstring>  
#include <cstdio>  
#include <queue>  
#include <stack>  
#include <algorithm>  
#include <cmath>  
using namespace std;
char dna[50][1010], ans[1010];
int c[5], h;
void Input(int m, int n)
{
    int i = 0, j = 0;
    for (i = 0; i < m; i++)
        for (j = 0; j < n; j++)
            cin >> dna[i][j];
}
void Judge(int j, int i)
{
    switch(dna[j][i])
    {
    case 'A': 
        c[0]++; break;
    case 'C': 
        c[1]++; break;
    case 'G':
        c[2]++; break;
    case 'T':
        c[3]++; break;
    }
}
void Count_H(int m, int n)
{
    int i = 0, j = 0;
    for (i = 0; i < m; i++)
        for (j = 0; j < n; j++)
            if (dna[i][j] != ans[j])
                h++;
}
int main()  
{  
#ifdef Local    
    freopen("a.in", "r", stdin);    
#endif
    int t = 0;
    cin >> t;
    while (t--)
    {
        h = 0;
        memset(dna, '\0', sizeof(dna));
        memset(ans, '\0', sizeof(ans));
        int m, n, i = 0, j = 0;
        cin >> m >> n;
        Input(m, n);
        for (i = 0; i < n; i++)
        {
            memset(c, 0, sizeof(c));
            for (j = 0; j < m; j++)
                Judge(j, i);
            int max = 0, pos = 0;
            for (j = 0; j < 4; j++)
            {
                if (c[j] > max)
                {
                    pos = j, max = c[j];
                }
            }
            switch(pos)
            {
            case 0:
                ans[i] = 'A'; break;
            case 1:
                ans[i] = 'C'; break;
            case 2:
                ans[i] = 'G'; break;
            case 3:
                ans[i] = 'T'; break;
            }
        }
        Count_H(m, n);
        cout << ans << endl;
        cout << h << endl;
    }
}