# POJ 3080 Blue Jeans （二分）

Blue Jeans

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 17537 Accepted: 7775

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

### AC 代码

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<set>
using namespace std;
typedef int LL;

char a[15][65];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d%*c",&n);
for(int i=0; i<n; i++)
gets(a[i]);
int low=1,high=60;      //初始子串最大长度与最小长度
string ans="";
while(low<=high)
{
int mid=(low+high)/2;   //二分
char str[mid+1];
bool flag=false;
set<string>sk;
for(int i=0; i<=60-mid; i++)    //枚举所有长为mid的子串，加入集合
{
strncpy(str,a[0]+i,mid);
str[mid]=0;
sk.insert((string)str);
}
for(set<string>::iterator j=sk.begin(); j!=sk.end(); j++)   //因为set默认已经排好序了，所以从字典序最小开始枚举
{
flag=false;
for(int i=1; i<n; i++)      //枚举其他字符串
if(!strstr(a[i],(*j).data()))   //若无这一个子串
{
flag=true;
break;
}
if(!flag)                   //若其他串都包含该子串
{
ans=*j;
break;
}
}
if(flag)high=mid-1;
else low=mid+1;
}
if(ans.size()<3)cout<<"no significant commonalities"<<endl;
else cout<<ans<<endl;
}
return 0;
}