DNA Consensus String UVA - 1368

喜欢ヅ旅行 2022-06-10 14:22 119阅读 0赞

## 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;
        }
    }