UVA_1368 - DNA Consensus String

Figure 1.

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.

 DNA sequence of gene XCat:GCATATGGCTGTGCADog:GCAAATGGCTGTGCAHorse:GCTAATGGGTGTCCACow:GCAAATGGCTGTGCAMonkey: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,...,s_m of lengthn , the consensus error between a string y of lengthn and the setS is the sum of the Hamming distances betweeny and eachs_i inS . If the consensus error betweeny and S is the minimum among all possible stringsy of lengthn ,y is called a consensus string ofS . 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 integersm

and n

which are separated by a single space. The integer m

(4m50)

represents the number of DNA sequences and n

(4n1000)

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

题意：

给定m个DNA序列，计算出一个DNA序列，此序列和给定的m个DNA序列的汉明码距离和最小，并求出该距离

解题思路：

读入m个序列到二维数组中，然后按列依次扫描，统计每一列碱基对A,C,G,T的个数，然后算出每一列A,C,G,T中个数最多的字母（注意ACGT按字母序排序），该字母就是应该输出的序列；同时计算每一列的hamming距离，最后累加即可。

代码如下：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int ROW = 50 + 5;const int COL = 1000 + 5;char dna[ROW][COL];char d[] = "ACGT";int cnt[4]; //统计一列A,C,G,T的个数int m,n;void read(){    int i,j;    for(i = 0;i<m;i++){        for(j = 0;j<n;j++)            cin>>dna[i][j];    }}char Search_cnt(int& hamming){    char c;    int pos;    int maxcnt = -1;    for(int i=0;i<4;i++){        if(cnt[i]>maxcnt) { maxcnt = cnt[i]; c = d[i]; pos = i; }    }    for(int i=0;i<4;i++){        if(i != pos&&cnt[i]!=0) hamming += cnt[i];    }    return c;}int main(){    //freopen("1368.txt","r",stdin);    int icase;cin>>icase;    while(icase--)    {        cin>>m>>n;        read();        string str;        int sum = 0;        int i,j;        for(j=0;j<n;j++)        {            memset(cnt,0,sizeof(cnt));            for(i=0;i<m;i++){                if(dna[i][j]=='A') cnt[0]++;                else if(dna[i][j]=='C') cnt[1]++;                else if(dna[i][j]=='G') cnt[2]++;                else if(dna[i][j]=='T') cnt[3]++;            }            int hamming = 0;            str += Search_cnt(hamming);            sum += hamming;        }        cout<<str<<endl<<sum<<endl;    }    return 0;}