f437. 1368 - DNA Consensus String
標籤 : CPE UVA 字串
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最近更新 : 2020-11-22 14:03

內容

給M個長度為N的DNA序列(由ATGC組成的字串)求出和這N個字串漢明距離最短的字串和其漢明距離

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  epresent 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.

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 onsensus 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.

輸入說明

第一行有一個T代表有幾組測資
每筆測資第一行有M,N(4 ≤ M ≤ 50)(4 ≤ N ≤ 1000)代表DNA數量和長度
接下來M行有長度為N的DNA序列

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.

輸出說明

輸出漢明距離最短的字串和其漢明距離
如果有多組解 取最小字典序

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.

範例輸入 #1
3
5 8
TATGATAC
TAAGCTAC
AAAGATCC
TGAGATAC
TAAGATGT
4 10
ACGTACGTAC
CCGTACGTAG
GCGTACGTAT
TCGTACGTAA
6 10
ATGTTACCAT
AAGTTACGAT
AACAAAGCAA
AAGTTACCTT
AAGTTACCAA
TACTTACCAA
範例輸出 #1
TAAGATAC
7
ACGTACGTAA
6
AAGTTACCAA
12
測資資訊:
記憶體限制: 512 MB
公開 測資點#0 (20%): 1.0s , <10M
公開 測資點#1 (20%): 1.0s , <10M
公開 測資點#2 (20%): 1.0s , <10M
公開 測資點#3 (20%): 1.0s , <10M
公開 測資點#4 (20%): 1.0s , <1M
提示 :

在資訊理論中,兩個等長字符串之間的漢明距離(英語:Hamming distance)是兩個字符串對應位置的不同字符的個數。換句話說,它就是將一個字符串變換成另外一個字符串所需要替換的字符個數。

以上出自維基百科

2020 6月CPE 第三題

標籤:
CPE UVA 字串
出處:
UVA1368 [管理者: DE45A (一葉之秋) ]

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