poj 3080 kmp求解多个字符串的最长公共字串，（数据小，有点小暴力 16ms）

Blue Jeans

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14113		Accepted: 6260

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<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
char test[][];
int next[];
int len1;
char head[];
 int n;
 int ma;
 char result[];
void getnext(){
    memset(next,,sizeof(next));
    int i=,j=-;
    next[]=-;
    while(i<len1){
        if(j==-||head[i]==head[j]){
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}

int  kmp(){
    ma=;
    for(int ti=;ti<n;ti++){
        int i=,j=,m=;
        while(i<){
            if(j==-||test[ti][i]==head[j]){
                i++;
                j++;
            }
            else
                j=next[j];
            if(j>m)
                m=j;
        }
        if(m<ma)
            ma=m;

    }
    return ma;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        memset(test,,sizeof(test));
        memset(result,,sizeof(result));
        scanf("%d",&n);
        getchar();
        for(int i=;i<n;i++){
            cin>>test[i];
        }
        int ans=;
        for(int i=;i<=;i++){
            memset(head,,sizeof(head));
            len1=-i;
            strcpy(head,test[]+i);   ///  枚举第一个串的所有后缀串（当然最后2个可以省去）
            head[len1]='\0';
            getnext();
            int temp=kmp(); ///  KMP求出这个后缀串与其余所有串的最大匹配。
            if(temp>ans){
                ans=temp;
                strncpy(result,test[]+i,ans);
            }
            else if(temp==ans){   ///  存在多个最长公共子串，输出字典序最小的，WA了一次。
                if(strcmp(test[]+i,result)<)
                    strncpy(result,test[]+i,ans);   ///  复习: strncpy()没有复制最后的'\0'。
            }

        }
        if(ans<)
            printf("no significant commonalities\n");
        else
            printf("%s\n",result);
    }
    return ;
}