POJ3080 Blue Jeans(暴力kmp)

Blue Jeans
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 18764 Accepted: 8358
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

暴力枚举第一个串的所有子串，然后让该子串与后面所有串匹配，找出能匹配所有串的最长子串，注意字典序最小

#include<stdio.h>#include<algorithm>#include<iostream>#include<string.h>#include<vector>using namespace std;int nxt[65];string s[15];void getnxt(string T){    nxt[0]=-1;    int i=0,j=-1;    int m=T.size();    while(i<m)    {        if(j==-1||T[i]==T[j])        {            nxt[++i]=++j;        }        else            j=nxt[j];    }}bool kmp(string S,string T){    int i=0,j=0;    int n=S.size();    int m=T.size();    while(i<n)    {        if(j==-1||S[i]==T[j])        {            i++;            j++;        }        else            j=nxt[j];   if(j==m)    return true;    }    return false;}int main(){    ios::sync_with_stdio(false);    cin.tie(0);    int t,n;    cin>>t;    while(t--)    {        cin>>n;        for(int i=0; i<n; i++)            cin>>s[i];        string ans="";        for(int i=1; i<=s[0].size(); i++)        {            for(int j=0; j<=s[0].size()-i; j++)            {               string S=s[0].substr(j,i);                bool flag=false;                getnxt(S);                for(int k=1; k<n; k++)                    {                        if(!kmp(s[k],S))                        {                            flag=true;                            break;                        }                    }                if(!flag)                {                    if(ans.size()<S.size())ans=S;                    else if(ans.size()==S.size())ans=min(ans,S);                }            }        }        if(ans.size()<3)cout<<"no significant commonalities"<<endl;        else            cout<<ans<<endl;    }    return 0;}