UVA - 1368 DNA Consensus String

最新推荐文章于 2025-08-13 23:15:00 发布

epsilon1

最新推荐文章于 2025-08-13 23:15:00 发布

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分类专栏：算法文章标签： c++ stl uva

本文链接：https://blog.youkuaiyun.com/qq_20480611/article/details/49177641

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本文介绍了一个算法，用于从多个DNA序列中找出每列出现次数最多的核苷酸，并生成共识序列及错误计数。该算法首先读取测试用例数量及每个案例的序列数量和长度，然后逐个比较并统计每种核苷酸的出现次数。

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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 $\le$ m $\le$ 50) represents the number of DNA sequences and n (4 $\le$ n $\le$ 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

分析：读取字符串，找到每列出现次数最多的字符。

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
using namespace std;
const int maxn = 1000 + 10;
int cnt[maxn][26];
char ans[maxn];

int main()
{
	int T;
	int m, n;
    string s;
	cin >> T;
	while(T--) {
		cin >> m >> n;
		memset(cnt, 0, sizeof(cnt));
		for(int i=0; i<m; i++) {
			cin >> s;
			for(int j=0; j<n; j++) {
				cnt[j][s[j]-'A']++;
			}
		}
		int diff = 0;
		for(int i=0; i<n; i++) {
			int total = 0, ch = 'A';
            for(int j=0; j<26; j++) {
				if(cnt[i][j]>total) {
					total = cnt[i][j];
					ch = 'A' + j;
				}
            }
            diff += m - total;
            ans[i] = ch;
		}
		for(int i=0; i<n; i++) putchar(ans[i]);
		printf("\n%d\n", diff);
	}
    return 0;
}