求最长连续公共子序列 POJ 3080

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




这个代码说明了STL真心强大，不过效率有点低~~~（比较的次数有点多）

#include<iostream>
#include<string.h>
#define MAXN 60
using namespace std;

int main()
{
int t;
cout<<"请输入测试次数"<<endl;
cin>>t; //测试数据的次数
while(t--)
{
int n; //每组测试数据有多少行数据
cout<<"请输入此数据的行数"<<endl;
cin>>n;
string str[n]; //用以存放数据
for(int i=0;i<n;i++)
{
cin>>str[i];
}
string res = " "; //用于存储找到的符合要求的子串
for(int i=3;i<=MAXN;i++) //i表示子串的长度
{
for(int j=0;j<MAXN;j++) //j表示字符串中的下标位置
{
string tmp=str[0].substr(j,i);
bool flag=true;
for(int k=1;k<n;k++)
{
if(str[k].find(tmp)==string::npos)
{
flag=false;
break;
}
}
if(flag&&res.size()<tmp.size()) res=tmp; //获得较长的子串
else if(flag&&(res.size()==tmp.size())&&(tmp<res)) res=tmp; //如果长度相等，则获得获得较小的字符串
}
}
if(res==" ") cout<<"no significant commonalities"<<endl;
else cout<<res<<endl;
}
return 0;
}

这道题跟算法中的求最长公共子序列有点像，不过区别是算法中的题的子串可以是不连续的，子要存在该字符串就可以了，采用的是递归回溯的方法，此处是连续的