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 |
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 (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
题目类型：字符处理题
算法分析：由本题的要求可知，只需对于每一个左到右，从上到下遍历已经存储在数组seq中的元素，将其与A、T、C、G顺序比较并算出每一列的Hamming距离sum，将所有列的Hamming距离都算出来后取其中的最小值并用temp记下最小值对应的A、T、C、G中的一个，累加到总Hamming距离ssum上并输出temp的值。以此类推，直到数组的最后一列之后，输出回车和ssum，下面第一个代码是更好的实现
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/************************************************** filename :m.cpp author :maksyuki created time :2017/12/2 12:45:27 last modified :2017/12/2 13:07:58 file location :C:\Users\abcd\Desktop\TheEternalPoet ***************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("in", "r", stdin) #define CFO freopen ("out", "w", stdout) #define CPPFF ifstream cin ("in") #define CPPFO ofstream cout ("out") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC) #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1666; const char v[] = {'*', 'A', 'C', 'G', 'T'}; int row, col; char g[maxn][maxn]; char s[maxn]; int check(char c1, char c2) { if(c1 == c2) return 0; return 1; } int main() { #ifdef LOCAL CFF; //CFO; #endif int t; scanf("%d", &t); while(t--) { scanf(" %d %d", &row, &col); for(int i = 1; i <= row; i++) for(int j = 1; j <= col; j++) scanf(" %c", &g[i][j]); int res = 0, len = 0; for(int i = 1; i <= col; i++) { int minval = INF; int pos = -1; for(int j = 1; j <= 4; j++) { int tmp = 0; for(int k = 1; k <= row; k++) tmp += check(v[j], g[k][i]); if(minval > tmp) { minval = tmp; pos = j; } } res += minval; s[len++] = v[pos]; } s[len] = 0; printf("%s\n%d\n", s, res); } return 0; } |
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#include <cstdio> #include <iostream> #include <fstream> using namespace std; const char flag[] = {'A', 'C', 'G', 'T'}; char seq[56][1006]; inline int absnum (int a) { return a >= 0 ? a : -a; } int main() { int cases, m, n; char temp; //ifstream cin ("aaa.txt"); cin >> cases; while (cases--) { cin >> m >> n; int i; for (i = 0; i < m; i++) cin >> seq[i]; int j, k, sum, ssum = 0; for (i = 0; i < n; i++) { int minval = 666666666; for (k = 0; k < 4; k++) { sum = 0; for (j = 0; j < m; j++) if (seq[j][i] != flag[k]) sum++; if (minval > sum) { minval = sum; temp = flag[k]; } } int q; for (q = 0; q < m; q++) if (seq[q][i] != temp) ssum++; printf ("%c", temp); } printf ("\n%d\n", ssum); } return 0; } |
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