POJ 3080 Blue Jeans(KMP 最長公用子串)

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Blue Jeans

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

32GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA3GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAGATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAAGATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA3CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output

no significant commonalitiesAGATACCATCATCAT

題意   給你n個DNA串   求它們的長度最大的公用子串   如果有多個  輸出字典序最小的   長度小於3的不算

每個DNA串的長度都是60  可以從子串長度為60依次遞減   並枚舉所有該長度子串  當某個長度的子串也為其它n-1個串的子串時   就是我們要的答案了

判斷是否為其它DNA串的子串直接kmp就行了

#include<cstdio>#include<cstring>using namespace std;const int N = 15, M = 65, L = 60;bool kmp (char s[], char p[]){    int next[M];    next[0] = -1, next[1] = 0;    int plen = strlen (p), i = 0, j,slen = strlen (s);    while (++i < plen - 1)    {        j = next[i];        while (j != -1 && p[i] != p[j])            j = next[j];        next[i + 1] = j + 1;    }    i = j = 0;    while (i < slen && j < plen)    {        if (j == -1 || s[i] == p[j]) ++i, ++j;        else j = next[j];    }    return j == plen;}int main(){    int cas, n, flag;    char s[N][M], tans[M],ans[M];    scanf ("%d", &cas);    while (cas--)    {        scanf ("%d", &n);        for (int i = flag = 0; i < n; ++i)            scanf ("%s", s[i]);        for (int lans = L; lans > 2 && flag == 0; --lans)        {            strcpy(ans,"ZZ");            for (int i = 0, j, k; i + lans <= L; ++i)            {                for (j = 0; j < lans; ++j)                    tans[j] = s[0][i + j];                tans[j] = '\0';                for (k = 1; k < n; ++k)                    if (!kmp (s[k], tans)) break;                if (k == n)                {                    if(strcmp(ans,tans)>0) strcpy(ans,tans);                    flag = 1;                }            }        }        if (flag == 1) printf ("%s\n", ans);        else printf ("no significant commonalities\n");    }    return 0;}