POJ 3080【KMP+暴力枚举】

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题目：

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

题目大意：

对于给定的n个定长的字符串，求解他们的最长公共子串。

解题方案：

想到了用暴力，，不过没敢用，简直了，，不过好歹用了一些KMP，很久之前看的东西，顺便整理一下。

解题代码：

 

#include<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1using namespace std;string s[N];
int Next[maxn];void getnext(string str,int slen)
{int k=-1;Next[0]=-1;for(int i=1;i<slen;i++){while(k>-1&&str[k+1]!=str[i])k=Next[k];if(str[k+1]==str[i])k++;Next[i]=k;}
}
bool kmp(string ptr,int plen,string str,int slen)   //判断母串中是否存在子串
{int k=-1;for(int i=0;i<plen;i++){while(k>-1&&str[k+1]!=ptr[i])k=Next[k];if(str[k+1]==ptr[i])k++;if(k==slen-1)return 1;}return 0;
}
int main()
{ios::sync_with_stdio(false);    //加速cin.tie(0);//   cout<<setiosflags(ios::fixed)<<setprecision(2);int t,n;cin>>t;while(t--){cin>>n;for(int i=0;i<n;i++)cin>>s[i];  //每组数据有n个字符串string ans="";for(int i=1;i<=s[0].size();i++)//长度{for(int j=0;j<=s[0].size()-i;j++)//起点{string op=s[0].substr(j,i);getnext(op,op.size());      //对于串1的子串求nextbool flag=0;for(int k=1;k<n;k++)        //判断剩余n-1个字符串中是否都存在该子串if(!kmp(s[k],s[k].size(),op,op.size()))flag=1;if(!flag)   //都存在该子串{if(ans.size()<op.size())ans=op;     //更新最长公共子串else if(ans.size()==op.size())ans=min(ans,op);}}}if(ans.size()<3)cout<<"no significant commonalities"<<endl;else cout<<ans<<endl;}return 0;
}

 

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