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  • POJ 3080, Blue Jeans

    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;
    }
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  • 原文地址:https://www.cnblogs.com/asuran/p/1577214.html
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