【KMP+二分枚举】 I - Blue Jeans POJ - 3080

Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 21193   Accepted: 9399 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 Source South Central USA 2006

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

char ch[15][70];
int nx[70];
char ch2[70], ans[70];

void cal_next(char b[], int m)
{
	nx[0] = -1;
	int k = -1;
	for (int i = 1; i < m; i++)
	{
		while (k > -1 && b[k + 1] != b[i])
			k = nx[k];
		if (b[k + 1] == b[i])
			k++;
		nx[i] = k;
	}
}

int KMP(char a[], char b[], int m)
{
	cal_next(b, m);

	int k = -1;
	for (int i = 0; i < 60; i++)
	{
		while (k > -1 && b[k + 1] != a[i])
			k = nx[k];
		if (b[k + 1] == a[i])
			k++;
		if (k == m - 1)
			return i - m + 2;
	}
	return -1;
}

int main()
{
	#ifndef ONLINE_JUDGE
		freopen("C:\\in.txt", "r", stdin);
	#endif // ONLINE_JUDGE

	int T;
	scanf("%d", &T);
	while (T--)
	{
		int n;
		scanf("%d", &n);
		for (int i = 0; i < n; i++)
			scanf("%s", ch[i]);

		int lans = -1;
		int l = 3, r = 60;
		while (l <= r)
		{
			int mid = (l + r) / 2;
			bool Flag = 0;

			for (int s = 0; s <= (59 - mid + 1) ; s++)
			{
				bool flag = 1;

				for (int l = 0; l < mid; l++)
					ch2[l] = ch[0][s + l];
				ch2[mid] = '\0';
				/// 从 ch[0] 中枚举取出长度为 mid 的字符串 存入 ch2

				/// 检查其余串中是否包含 ch2[]
				for (int i = 1; i < n; i++)
				{
					int temp = KMP(ch[i], ch2, mid);
					if (temp == -1)  // 有不包含的
						flag = 0;
					if (!flag)
						break;
				}

				if (flag) // 都包含
				{
					if (mid > lans)
						strcpy(ans, ch2);
					else if (mid == lans && strcmp(ch2, ans) < 0) // ch2 < ans
						strcpy(ans, ch2);

					Flag = 1; // 当前mid长度可取
					lans = mid;
				}
			}

			if (Flag) // mid长度可取
				l = mid + 1;
			else
				r = mid - 1;
		}

		if (lans == -1)
			printf("no significant commonalities\n");
		else
			printf("%s\n", ans);
	}

	return 0;
}