POJ 3080, Blue Jeans

转载于 2019-06-29 11:57:09 发布 · 219 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/asuran/archive/2009/10/01/1577214.html

本文介绍了一种用于比对多个DNA序列并找出最长公共子序列的算法，此算法通过逐个检查所有可能的子串来确定至少三个碱基长度的共同序列。在给定的DNA片段中，该算法能有效找出相关联的个体调查信息，识别新的遗传标记。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

brute force

用第一个字符串从长到短生成所有的子串，并和其他字符串匹配。

Description
The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated.

As an IBM researcher, you have been tasked with writing a program that will find commonalities amongst given snippets of DNA that can be correlated with individual survey information to identify new genetic markers.

A DNA base sequence is noted by listing the nitrogen bases in the order in which they are found in the molecule. There are four bases: adenine (A), thymine (T), guanine (G), and cytosine (C). A 6-base DNA sequence could be represented as TAGACC.

Given a set of DNA base sequences, determine the longest series of bases that occurs in all of the sequences.

Input
Input to this problem will begin with a line containing a single integer n indicating the number of datasets. Each dataset consists of the following components:
A single positive integer m (2 <= m <= 10) indicating the number of base sequences in this dataset.
m lines each containing a single base sequence consisting of 60 bases.

Output
For each dataset in the input, output the longest base subsequence common to all of the given base sequences. If the longest common subsequence is less than three bases in length, display the string "no significant commonalities" instead. If multiple subsequences of the same longest length exist, output only the subsequence that comes first in alphabetical order.

Sample Input
3
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output
no significant commonalities
AGATAC
CATCATCAT

Source
South Central USA 2006

// POJ3080.cpp : Defines the entry point for the console application.
//

#include <iostream>
#include <string>
using namespace std;

string ComSubStr(string* ln, int n)
{
    string longest="";
    for (int j = 60; j > 2; --j)
        for (int i = 0; i + j <= 60; ++i)
        {
            string str = ln[0].substr(i, j);
            bool find = true;
            for (int k = 1; k < n; ++k)
                if (ln[k].find(str) == string::npos)
                {
                    find = false;
                    break;
                };

            if (find == true && str.size() > longest.size())longest = str;
            else if(find == true && str.size() == longest.size() && str < longest) longest = str;
        };
    return longest;
};
int main(int argc, char* argv[])
{
    int cases;
    cin >> cases;

    int lines;
    string ln[10];
    for (int c = 0; c < cases; ++c)
    {
        cin >> lines;
        for (int i = 0; i < lines; ++i)
        {
            cin >> ln[i];
        }

        string r = ComSubStr(ln, lines);
        if (r == "") cout << "no significant commonalities\n";
        else cout << r << endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/asuran/archive/2009/10/01/1577214.html