poj 3080 Blue Jeans KMP多模式匹配

A - Blue Jeans(KMP)

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 3080

Appoint description:  System Crawler  (2016-01-21)

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

题目大意：找多个字符串中的最长公共连续子串，如果长度相同，则选取字典序最小的

解题思路：暴力大法。首先找出最短的串，枚举这个串的所有子串，让其与每一个字符串进行匹配，如果匹配成功，则说明这个子串是一个可能的解。

优化策略：首先，匹配的时候用KMP，其次，枚举子串先枚举最长的，这样一旦找到就不用枚举更短的了，比较字典序可以用strcmp。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#define MAX 70
using namespace std;

int Next[MAX];
char str[11][70];
char ans[MAX];

void get_next(char x[],int m,int Next[])
{
    int i,j;
    j=Next[0]=-1;
    i=0;
    while(i<m)
    {
        while(-1!=j&&x[i]!=x[j]) j=Next[j];
        if(x[++i]==x[++j]) Next[i]=Next[j];
        else Next[i]=j;
    }
}

bool kmp(char x[],int m,char y[],int n)
{
    int i,j;
    get_next(x,m,Next);
    i=j=0;
    while(i<n)
    {
        while(-1!=j&&y[i]!=x[j]) j=Next[j];
        i++;
        j++;
        if(j>=m)
            return true;
    }
    return false;
}







int main()
{
    int t,n,min_l,min_p;
    int i,j,k,l;
    bool flag;
    char tmp[70];
    scanf("%d",&t);
    while(t--)
    {
        flag=false;
        min_l=MAX;
        min_p=0;
        scanf("%d",&n);
        for(i=0; i<n; i++)
        {
            scanf("%s",str[i]);
            if(min_l>strlen(str[i]))
            {
                min_l=strlen(str[i]);
                min_p=i;
            }
        }
        for(i=min_l; i>=3&&flag==false; i--) //枚举长度
        {
            for(j=0; j<=min_l-i; j++) //枚举子串
            {
                for(l=j; l<i+j; l++)
                    tmp[l-j]=str[min_p][l];
                tmp[l]='\0';
                for(k=0; k<n; k++) //逐个匹配
                {
                    if(k==min_p)
                        continue;
                    if(kmp(tmp,i,str[k],strlen(str[k]))==false)
                        break;
                }
                if(k>=n)
                {
                    if(flag==false)
                    {
                        flag=true;
                        strcpy(ans,tmp);
                    }
                    else
                    {
                        if(strcmp(ans,tmp)>0)
                            strcpy(ans,tmp);
                    }
                }


            }
        }
        if(flag==true)
            printf("%s\n",ans);
        else
            printf("no significant commonalities\n");
    }
    return 0;
}