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Blue Jeans
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 21051 | Accepted: 9323 |
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
另外在提供一些测试样例：
输入：
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 输出： no significant commonalities AGATAC CATCATCAT TGAT GAT no significant commonalities AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA no significant commonalities AAA ACTACT
题意：寻找n个60位DNA字符串的共同子序列。
题解：首选想到的当然是暴力了。找出一个DNA的所以子串，然后判断是否是其他DNA的子串。算一下时间复杂度，因为DNA固定长度为60，求出所以子串的复杂度也就是60*60，然后再去判断其他的子串，用kmp的时间复杂度为60+60，总的时间复杂度为60*60*（60+60）*10，这样看来也不是很大，果断暴力+kmp。
ps：题目要求如果有多个长度一样的子序列，则按字典序输出第一个。代码里我用了STL使代码看起来简单一点。
具体代码如下：
#include<iostream>
#include<string>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
int len=60;
int kmp(string a,string b)
{
int i=-1,j=0;
int next[61];
next[0]=-1;
int blen=b.size();
while(j<blen-1)
if(i==-1||b[j]==b[i])
if(next[++i]==next[++j])
next[j]=next[i];
else
next[j]=i;
else
i=next[i];
i=j=0;
while(i<len&&j<blen)
if(j==-1||a[i]==b[j])
i++,j++;
else
j=next[j];
if(j==blen)
return 1;
else
return -1;
}
int main()
{
int T;
cin>>T;
while(T--)
{
int n;
cin>>n;
string str;//作为模板
cin>>str;
vector<string> s;
priority_queue<string> ans;
ans.push("1");
for(int i=0;i<n-1;i++)
{
string a;
cin>>a;
s.push_back(a);
}
for(int j=0;j<len-2;j++)
{
for(int k=3;k<=len-j;k++)
{
string s1(str,j,k);//找到子序列存进s1
int cnt=0;
for(int i=0;i<n-1;i++)
{
if(kmp(s[i],s1)==-1)
break;
else//找到了
cnt++;
}
if(cnt==n-1)
{
string old=ans.top();
int LL=old.size();
if(s1.size()>old.size())
{
ans.pop();
ans.push(s1);
}
else if(s1.size()==LL)
ans.push(s1);
}
}
}
if(ans.top()=="1")
cout<<"no significant commonalities"<<endl;
else
{
while(ans.size()>2)
{
ans.pop();
}
string ans1=ans.top();
ans.pop();
string ans2=ans.top();
int ans1len=ans1.size();
int ans2len=ans2.size();
if(ans1len>ans2len)
cout<<ans1<<endl;
else
cout<<ans2<<endl;
}
}
return 0;
}