3080 Blue Jeans 求N个字符串的最长子串 暴力枚举+ KMP模板

Blue Jeans

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4913		Accepted: 2033

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

#include
#include
#include
using namespace std;
int next[100];
void get_next(char *s,int next[])
{
    int len=strlen(s);
    int i=0,j=-1;
    next[0]=-1;
    while(i    {
        if(j==-1||s[i]==s[j]) i++,j++,next[i]=j;
        else j=next[j];
    }
}
int KMP(char *a,char *b)//a主串，b子串 返回值为a中与b匹配的第一个字母的位置(从0开始)
{
    get_next(b,next);//计算next
    int n=strlen(a),m=strlen(b);
    int i=0,j=0;
    while(i    {
        if(j==-1||a[i]==b[j]) i++,j++;
        else j=next[j];
    }
    if(j>=m) return i-m;
    return -1;
}
int KMP_Count(char *a,char *b)//a主串，b子串 返回值为共有a中多少个与b匹配的子串
{
    get_next(b,next);//计算next
    int n=strlen(a),m=strlen(b);
    int i=0,j=0;
    int cnt=0;
    while(i    {
        if(j==-1||a[i]==b[j]) i++,j++;
        else j=next[j];
        if(j==m) cnt++,j=next[j];
    }
    return cnt;
}
char map[100][100];
char s[100],_max[100],res;
int main()
{
    int ci;scanf("%d",&ci);
    while(ci--)
    {
        _max[0]='/0';res=0;
        int n;scanf("%d",&n);
        for(int i=0;i        for(int i=0,len=strlen(map[0]);i        {
            for(int j=i;j            {
                int l=0;
                for(int k=i;k<=j;k++) s[l++]=map[0][k];s[l]='/0';
                int flag=1;
                for(int i=1;i                {
                    if(KMP(map[i],s)==-1)
                    {
                        flag=0;break;
                    }
                }
                if(flag)
                {
                    if(l>res) res=l,strcpy(_max,s);
                    else if(l==res)
                    {
                        if(strcmp(s,_max)<0) strcpy(_max,s);
                    }
                }
            }
        }
        if(res<3) printf("no significant commonalities/n");
        else printf("%s/n",_max);
    }
    return 0;
}