uvaoj-1368:DNA序列
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 X | |
| Cat: | GCATATGGCTGTGCA |
| Dog: | GCAAATGGCTGTGCA |
| Horse: | GCTAATGGGTGTCCA |
| Cow: | GCAAATGGCTGTGCA |
| Monkey: | GCAAATCGGTGAGCA |
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 stringsS =s1 ,...,sm of lengthn, the consensus error between a stringy of lengthn and the setS is the sum of the Hamming distances betweeny and eachsi inS. If the consensus error betweeny andSis 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 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
题解：简单的贪心的题，通过对每一位的判断来选出当前位置次数最高的DNA脱氧核糖核苷酸；
code：
#include
#include
using namespace std;
int main()
{
int T;
int m,n;
char dna[55][1050];
char ans[1050];//最终题目求得答案序列；
int a,t,g,c;//用来存当前位置每一种碱基的数目；
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&m,&n);
for(int i=0; i
scanf("%s",dna[i]);
}
for(int i=0; i
int maxn=0;//每一次都需要初始化；
a=0;
t=0;
g=0;
c=0;
for(int j=0; j
if(dna[j][i]=='A')//判断每一位的种类，用switch语句更好一点；
{
a++;
}
if(dna[j][i]=='C')
{
c++;
}
if(dna[j][i]=='G')
{
g++;
}
if(dna[j][i]=='T')
{
t++;
}
}
if(a>maxn)//在判断的同时保证了字典序；
{
maxn=a;
ans[i]='A';
}
if(c>maxn)
{
maxn=c;
ans[i]='C';
}
if(g>maxn)
{
maxn=g;
ans[i]='G';
}
if(t>maxn)
{
maxn=t;
ans[i]='T';
}
}
ans[n]='\0';
//
int num=0;
for(int i=0; i
for(int j=0; j
if(ans[j]!=dna[i][j])
{
num++;
}
}
}
puts(ans);
printf("%d\n",num);
}
return 0;
}