7-12 碱基序列匹配 （25 分）
碱基序列匹配
地理项目是IBM和国家地理学会的合作研究项目，从成千上万捐献的DNA分析地球上人类是如何繁衍的。作为一个IBM的研究人员，请你写一个程序找出给定的DNA片段之间的相同之处，使得对个体的调查相关联。一个DNA碱基序列是指把在分子中发现的氮基的序列给罗列出来。有四种氮基：腺嘌呤 (A)、胸腺嘧啶(T)、鸟嘌呤(G)和胞嘧啶(D)，例如，一个6碱基DNA序列可以表示为 TAGACC。给出一个DNA碱基序列的集合，确定在所有序列中都出现的最长的碱基序列。
输入格式:
输入的第一行给出了整数n，表示测试数据集合的数目。每个测试数据集合由下述两部分组成：一个正整数m(2≤m≤10)，给出数据集合中碱基序列的数目。m行，每行给出一个60碱基的碱基序列。
输出格式:
对于输入的每个测试数据集合的所有的碱基序列，输出最长的相同的碱基子序列。如果最长的相同的碱基子序列的长度小于3，则输出“no significant commonalities”来代替碱基子序列。如果相同最长长度的子序列有多个，则仅输出按字母排序的第一个。
输入样例1:
3
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
输出样例1:
no significant commonalities
AGATAC
CATCATCAT
输入样例2:
5
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
2
GATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGAT
GATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATT
10
GATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGAT
GATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATT
GATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTT
GATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTT
AGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGAT
AAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGAT
AAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGAT
CGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGAT
CCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGAT
CCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGAT
输出样例2:
no significant commonalities
AGATAC
CATCATCAT
TGAT
GAT
输入样例3:
11
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
2
GATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGAT
GATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATT
10
GATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGAT
GATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATT
GATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTT
GATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTT
AGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGAT
AAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGAT
AAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGAT
CGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGAT
CCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGAT
CCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGAT
10
GATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATC
GCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTA
TAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGC
CGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGAT
GTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGT
GAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGA
GCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGC
ATATATATATATATATATATATATATATATATATATATATATATATATATATATATATAT
ACACACACACACACACACACACACACACACACACACACACACACACACACACACACACAC
TCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTC
10
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
10
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAG
GGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAG
GGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGG
GGGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGG
GGGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGGG
GGGGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGGG
GGGGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGGGG
GGGGGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGGGG
4
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG
TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
10
AAAAAATAAAAAATAAAAAATAAAAAATAAAAAATAAAAAATAAAAAATAAAAAATAAAA
AAAAACAAAAACAAAAACAAAAACAAAAACAAAAACAAAAACAAAAACAAAAACAAAAAC
AAAAAGAAAAAGAAAAAGAAAAAGAAAAAGAAAAAGAAAAAGAAAAAGAAAAAGAAAAAG
AAAAATAAAAATAAAAATAAAAATAAAAATAAAAATAAAAATAAAAATAAAAATAAAAAT
AAAACAAAACAAAACAAAACAAAACAAAACAAAACAAAACAAAACAAAACAAAACAAAAC
AAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAG
AAAATAAAATAAAATAAAATAAAATAAAATAAAATAAAATAAAATAAAATAAAATAAAAT
AAACAAACAAACAAACAAACAAACAAACAAACAAACAAACAAACAAACAAACAAACAAAC
AAATAAATAAATAAATAAATAAATAAATAAATAAATAAATAAATAAATAAATAAATAAAT
AAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAG
2
GATGATGCATCATGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGACTACTAA
GATGATCATCATACTACTCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACTACTAGATGATGCATCATGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGA
ACTACTCGATGATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
输出样例3:
扫描二维码关注公众号，回复：
4593320 查看本文章
no significant commonalities
AGATAC
CATCATCAT
TGAT
GAT
no significant commonalities
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
no significant commonalities
AAA
ACTACT
此题博主是用字符串和KMP算法及遍历每一个子串做出来的。
代码如下：
#include<bits/stdc++.h>
using namespace std;
void getNext(string S,int* next) //得到子串下面的数组
{
int j,k;
j=0;k=-1;
next[0]=-1; //子串0号元素下面数为-1
while(j<(S.size()-1)) //对子串所有元素下面赋值
{
if(k==-1||S[j]==S[k]) //如果k回到了第一个元素或者第j个元素等于第k个元素
{
j++;k++; //j++；k++；
next[j]=k; //子串第j个元素下面的数为k
}
else
k=next[k]; //k为第子串第k个元素下面的数
}
}
bool bijiao(string T,string *S,int n) //返回该子串是否是每一个序列的子串
{
int a=T.size(); //得到子串T的长度
int next[a]; //建立子串的数组下标
getNext(T,next); //给子串数组赋值
int hhh[n]; //建立一个大小为n的数组判断子串是不是n-1个主串的公共子串
for(int i=0;i<n;i++)
{
hhh[i]=0; //给数组hhh全赋初值0
}
for(int l=1;l<n;l++)
{
int aa=S[l].size(); //得到主串的长度
int i=0,j=0;
while(i<aa) //当主串下标没到达尾部时
{
if(j==-1||S[l][i]==T[j]){
++i;
++j;
}
else
j=next[j];
if(j==a){
hhh[l]=1;
break;
}
}
}
for(int i=1;i<n;i++) //查看该子串是否为每一个主串下面的子串
{
if(hhh[i]!=1)
return false; //不是则返回false
}
return true; //反之是则返回true
}
void chuli(string *aar,int n)
{
string key="Z"; //假定最长公共序列key
string try1; //第一个碱基序列的每一个字串
int w=0;
for(int i=60;i>=3;i--) //从最长字符串长度开始作为子串长度
{
if(w!=0&&i<w)
{
cout<<key<<endl;return ;
}
for(int k=0;k<=60-i;k++) //开始位置
{
try1=aar[0].substr(k,i); //第一个碱基序列的一个字串
//cout<<try1<<endl;
if(bijiao(try1,aar,n)) //查看是否为公共字串
{
w=i;
if((try1.size()>=key.size())&&(try1<key))
key=try1;
}
}
}
if(key.size()<3)
cout<<"no significant commonalities"<<endl;
else
cout<<key<<endl;
}
int main()
{
int N;
cin>>N; //输入数据集合的数目N
for(int z=0;z<N;z++) //输入集合的每一组元素
{
int n;
cin>>n; //输入数据集合中碱基序列的数目n
string jjsz[n]; //建立jjsz[n]数组存放每一组碱基序列
for(int x=0;x<n;x++)
{
cin>>jjsz[x]; //存放每一个碱基序列
}
chuli(jjsz,n); //开始处理
}
return 0;
}