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本文链接：https://blog.youkuaiyun.com/chAngE_AC/article/details/47403169

本文介绍了一种用于比对多个固定长度DNA序列的算法，旨在找出最长的共同子序列。通过对给定样本进行分析，该算法能够有效识别出至少三个碱基长度的共有片段，并在存在多个等长片段时选择字典序最小的一个。

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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

题意：给你几个DNA序列长度为60，以第一个为模板，找到之后的DNA中与模板DNA相同的子序列，且保证子序列最长（长度大于等于3）。

依旧是暴力找就行，枚举相同序列的长度以第一个DNA为模板向其他串中找。其中有个技巧性的地方就是strstr（）函数的使用，strstr(a,b)函数为在a中找b，如果可以找到b那么会返回最初始找到b时的位置的地址，若找不到b则返回NULL。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#define INF 0x3f3f3f3f

using namespace std;

int main()
{
    int t,n;
    int i,j,k;
    char str[11][65],ans[65],strm[65];
    scanf("%d",&t);
    while (t--)
    {
        ans[0]='\0';
        scanf("%d%*c",&n);
        for (i=0;i<n;i++)
        {
            scanf("%s",str[i]);
        }
        for (int len=3;len<=60;len++)
        {
            for (j=0;j<=60-len;j++)
            {
                strm[0]='\0';
                for (k=j;k<j+len;k++)
                {
                    strm[k-j]=str[0][k];
                }
                strm[len]='\0';
                for (i=1;i<n;i++)
                {
                    char *p=strstr(str[i],strm);
                    if(p==NULL) break;
                }
                if(i==n)
                {
                    if(ans[0]=='\0') strcpy(ans,strm);
                    else if(strlen(ans)==strlen(strm))
                    {
                        if(strcmp(ans,strm)>0)
                            strcpy(ans,strm);
                    }
                    else if(strlen(ans)<strlen(strm))
                    {
                        strcpy(ans,strm);
                    }
                }
            }
        }
        if(ans[0]=='\0') printf("no significant commonalities\n");
        else
        printf("%s\n",ans);
    }
    return 0;
}