【白书之路】 1368 - DNA Consensus String 最相似串

DNA Consensus String

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-Cytosine-Cytosine-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.
3602 - DNA Consensus String 1/3
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. The following shows sample input and output for three test cases.
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

题目大意：寻找一个字符串，使其与一组字符串的相似度最高，对于任意两个串，他们的相异性定义为他们中相同位置不同字符的个数，相似度最高要求该串与每个串的相异性之和最小，如果有多个串，找到字典序最小的。

对于所求字符串的每一个位置，我们不难得到，它一定要取 在这个位置上出现频率最高的字符，即每一列上出现频率最高的字符。

如果字符出现的频率最高是1，那么这个位置不论取什么字符，都不会对相似度造成影响，因为题目中字符串的数量是大于4的,因此就取字典序最小的字符‘A'，否则取出现频率最高的字符。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
#include <memory.h>
#define MAX_LENGTH 1010
#define MAX_NUM 60
using namespace std;

char dna[MAX_NUM][MAX_LENGTH];//原始字符数组
int word_count[4];//四种字符出现频率统计
char res[MAX_LENGTH];//存放结果字符串
int m,n,t;
map<char,int> get_pos;//由字符获得坐标
map<int,char> get_char;//由坐标获得字符



void init()//初始化索引
{
    get_pos['A']=0;
    get_pos['C']=1;
    get_pos['G']=2;
    get_pos['T']=3;
    get_char[0]='A';
    get_char[1]='C';
    get_char[2]='G';
    get_char[3]='T';
}

int get_dis()//获得结果与每个字符串的相异度
{
    int sum=0;
    for(int i=0;i<m;i++)
    {
        for(int j=0;j<n;j++)
        {
            if(res[j]!=dna[i][j])
                sum++;
        }
    }
    return sum;
}

void get_con()//获取相似度最大的串
{
    int dis;
    int r,c;
    int max_pos;
    int max_count;
    for(c=0;c<n;c++)
    {
        memset(word_count,0,sizeof(word_count));
        for(r=0;r<m;r++)//统计
        {
            word_count[get_pos[dna[r][c]]]++;
        }
        max_count=word_count[0];//寻找最大频率字符
        max_pos=0;
        for(int i=1;i<4;i++)
        if(max_count<word_count[i])
        {
            max_count=word_count[i];
            max_pos=i;
        }
        if(max_count==1)//如果最大频率为1
            res[c]='A';
        else
            res[c]=get_char[max_pos];

    }
    res[c]='\0';//别忘了字符串结尾的'\0'
    dis=get_dis();
    printf("%s\n%d\n",res,dis);
}

int main()
{

    int i;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&n);
        for(i=0; i<m; i++)
        {
            scanf("%s",dna[i]);
        }
        get_con();
    }
    return 0;
}