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

Source

題意：求N個字串的最長公共子串 思路，直接把POJ3450的程式碼拿過來就行了

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 1000005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
int wa[N],wb[N],wsf[N],wv[N],sa[N];
int rank[N],height[N],s[N];
//sa:字典序中排第i位的起始位置在str中第sa[i]
//rank:就是str第i個位置的字尾是在字典序排第幾
//height：字典序排i和i-1的字尾的最長公共字首
int cmp(int *r,int a,int b,int k)
{
    return r[a]==r[b]&&r[a+k]==r[b+k];
}
void getsa(int *r,int *sa,int n,int m)//n要包含末尾新增的0
{
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0; i<m; i++)  wsf[i]=0;
    for(i=0; i<n; i++)  wsf[x[i]=r[i]]++;
    for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
    for(i=n-1; i>=0; i--)  sa[--wsf[x[i]]]=i;
    p=1;
    j=1;
    for(; p<n; j*=2,m=p)
    {
        for(p=0,i=n-j; i<n; i++)  y[p++]=i;
        for(i=0; i<n; i++)  if(sa[i]>=j)  y[p++]=sa[i]-j;
        for(i=0; i<n; i++)  wv[i]=x[y[i]];
        for(i=0; i<m; i++)  wsf[i]=0;
        for(i=0; i<n; i++)  wsf[wv[i]]++;
        for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
        for(i=n-1; i>=0; i--)  sa[--wsf[wv[i]]]=y[i];
        t=x;
        x=y;
        y=t;
        x[sa[0]]=0;
        for(p=1,i=1; i<n; i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
    }
}
void getheight(int *r,int n)//n不儲存最後的0
{
    int i,j,k=0;
    for(i=1; i<=n; i++)  rank[sa[i]]=i;
    for(i=0; i<n; i++)
    {
        if(k)
            k--;
        else
            k=0;
        j=sa[rank[i]-1];
        while(r[i+k]==r[j+k])
            k++;
        height[rank[i]]=k;
    }
}

char str[N],ans[N];
int id[N],vis[4005];

bool check(int mid,int n,int k)
{
    int i,j,cnt = 0;
    MEM(vis,0);
    for(i = 2; i<=n; i++)
    {
        if(height[i]<mid)
        {
            MEM(vis,0);
            cnt = 0;
            continue;
        }
        if(!vis[id[sa[i-1]]])
        {
            cnt++;
            vis[id[sa[i-1]]] = 1;
        }
        if(!vis[id[sa[i]]])
        {
            cnt++;
            vis[id[sa[i]]] = 1;
        }
        if(cnt == k)
        {
            for(j = 0; j<mid; j++)
                ans[j] = s[sa[i]+j];
            ans[mid] = '\0';
            return 1;
        }
    }
    return 0;
}

int main()
{
    int n,i,j,k,len,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&k);
        n = 0;
        for(i = 0; i<k; i++)
        {
            scanf("%s",str);
            len = strlen(str);
            for(j = 0; j<len; j++)
            {
                s[n] = str[j];
                id[n] = i;
                n++;
            }
            s[n] = '#'+i;
            id[n] = '#'+i;
            n++;
        }
        s[n] = 0;
        getsa(s,sa,n+1,5000);
        getheight(s,n);
        int l = 1,r = len,mid,flag = 0;
        while(l<=r)
        {
            mid = (l+r)/2;
            if(check(mid,n,k))
            {
                flag = mid;
                l=mid+1;
            }
            else
                r=mid-1;
        }
        if(flag>=3)
            printf("%s\n",ans);
        else
            printf("no significant commonalities\n");
    }

    return 0;
}