poj 3080 Blue Jeans【KMP】【求公共子串】

Blue Jeans

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15357		Accepted: 6823

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

此题同  poj3450  只不过最后计算出的公共子串的长度要不小于3！

AC代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#define mem(a, b) memset(a, (b), sizeof(a))
#define Wi(a) while(a--)
#define Si(a) scanf("%d", &a)
#define Pi(a) printf("%d\n", (a))
#define INF 0x3f3f3f3f
#include<algorithm>
#define N 100000+10
using namespace std;
char str[11][70];
int p[70];
void getp(char *s)
{
	int len = strlen(s);
	p[0] = p[1] = 0;
	for(int i = 1; i < len; i++)
	{
		int j = p[i];
		while(j && s[i]!=s[j])  j = p[j];
		p[i+1] = s[i]==s[j] ? j+1 : 0;
	}
}
bool find(char *a, char  *b)
{
	int la = strlen(a);
	int lb = strlen(b);
	int j = 0;
	for(int i = 0; i < la; i++)
	{
		while(j && a[i]!=b[j])  j = p[j];
		if(a[i] == b[j])  j++;
		if(j >= lb)  return true; 
	}
	return false;
}
int main()
{
	int t; Si(t);
	Wi(t)
	{
		int n; Si(n);
		for(int i = 0; i < n; i++)  scanf("%s", str[i]);
		int l = strlen(str[0]);
		char ch[70], ans[70];
		mem(ans, '\0');
		for(int i = 0; i < l; i++)
		{
			for(int j = i; j < l; j++)
			{
				int m = 0;
				for(int k = i; k <= j; k++)
					ch[m++] = str[0][k];
				ch[m] = '\0';
				getp(ch);
				bool flag = 1;
				for(int k = 1; k < n; k++)
				{
					if(!find(str[k], ch))
					{
						flag = 0;  break;
					}
				}
				if(flag)
				{
					if(strlen(ch) > strlen(ans))
						strcpy(ans, ch);
					else if(strlen(ch)==strlen(ans) && strcmp(ch, ans) < 0)
						strcpy(ans, ch);	
				}
			}
		}
		if(strlen(ans) < 3)   printf("no significant commonalities\n");
		else  printf("%s\n", ans);
	}
	return 0;
}