POJ 3080 Blue Jeans（字符串处理函数）

Blue Jeans

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

• 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

点击打开链接

#include <iostream>
#include<algorithm>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <string>

using namespace std;

string a[11];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=0; i<n; i++)
        {
            cin>>a[i];
        }
        string ans="";
        int ll = a[0].size();
        for(int i=0; i<ll-2; i++)
        {
            for(int j=3; j<=ll; j++)
            {
                string ss=a[0].substr(i,j);  ///自i起截取j个字符
                bool flag=true;
                for(int k=1; k<n; k++)
                {
                    if(a[k].find(ss)==string::npos)   /// 表示在a[k]中找不到
                    {
                        flag=false;
                        break;
                    }
                }
                if(flag && ans.length() < ss.length())
                {
                    ans=ss;
                }
                else if(flag && ans.length() == ss.length() && ans > ss)
                {
                    ans=ss;
                }
            }
        }
        if(ans == "")
        {
            printf("no significant commonalities\n");
        }
        else
        {
            cout << ans << endl;
        }
    }
    return 0;
}