UVA 1368 - DNA Consensus String(贪心)

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

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 stringsS=s₁,...,s_mof lengthn, the consensus error between a stringyof lengthnand the setSis the sum of the Hamming distances betweenyand eachs_iinS. If the consensus error betweenyandSis the minimum among all possible stringsyof lengthn,yis 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 ofTtest cases. The number of test casesTis given in the first line of the input. Each test case starts with a line containing two integersmandnwhich are separated by a single space. The integerm(4m50)represents the number of DNA sequences andn(4n1000)represents the length of the DNA sequences, respectively. In each of the nextmlines, 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n个DNA，求一个目标串，使得每个序列变换成目标所需次数最少。

思路：贪心。每一位选存在最多的变化即可。

代码:

#include <stdio.h>
#include <string.h>
const int N = 1005;
const int M = 55;
int t, n, m, vis[500], ans;
char g[M][N];

void init() {
    ans = 0;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i ++)
	scanf("%s", g[i]);
}

char cal(int i) {
    char c; int Max = 0; memset(vis, 0, sizeof(vis));
    for (int j = 0; j < n; j ++) {
	vis[g[j][i]] ++;
	if (Max < vis[g[j][i]]) {
	    Max = vis[g[j][i]];
	    c = g[j][i];
	}
	else if (Max == vis[g[j][i]] && g[j][i] < c) {
	    c = g[j][i];
	}
    }
    ans += (n - vis[c]);
    return c;
}

void solve() {
    for (int i = 0; i < m; i ++)
	printf("%c", cal(i));
    printf("\n%d\n", ans);
}

int main() {
    scanf("%d", &t);
    while (t--) {
	init();
	solve();
    }
    return 0;
}