Description
Input
Output
Sample Input
3
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
Sample Output
no significant commonalities
AGATAC
CATCATCAT
题意 给你n个DNA串 求它们的长度最大的公共子串 如果有多个 输出字典序最小的 长度小于3的不算
每个DNA串的长度都是60 可以从子串长度为60依次递减 并枚举所有该长度子串 当某个长度的子串也为其它n-1个串的子串时 就是我们要的答案了
判断是否为其它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;
}
POJ 3080 Blue Jeans(KMP 最长公共子串),布布扣,bubuko.com
POJ 3080 Blue Jeans(KMP 最长公共子串)
原文地址：http://blog.csdn.net/iooden/article/details/38457875
