(kmp) Blue Jeans (3080)

最新推荐文章于 2020-07-15 10:41:37 发布

原创最新推荐文章于 2020-07-15 10:41:37 发布 · 583 阅读

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本文探讨如何通过编程解决复杂DNA序列问题，旨在识别不同样本间的共同遗传特征。利用KMP算法处理字符串匹配，高效找出最长公共子序列，应用于基因组研究中识别遗传标记。

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这个题，感觉有一点复杂，中途有一点想放弃，，还是坚持下来了。还一次成功了。不错。，，今天写kmp比较顺利，也算是更理解一些了吧

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<set>
#include<cstring>

using namespace std;

int n;
char s[11][66];
int next[66];

int main()
{
	//freopen("in.txt","r",stdin);
	int i,j,k;

	int t;
	cin>>t;
	while (t--)
	{
		cin>>n;
		for (i=0;i<n;i++)
			cin>>s[i];

		int ans=0;
		char cur[66];
		char c[66];
		for (i=0;i<60;i++)
		{
			if (60-i<ans)
				break;
			k=0;
			for (j=i;j<60;j++,k++)
			{
				cur[k]=s[0][j];
			}
			
			int num=0;
			j=-1;
			next[0]=-1;
			while (num<k)
			{
				if (j==-1||cur[j]==cur[num])
				{
					num++;
					j++;
					next[num]=j;
				}
				else
					j=next[j];
			}
			int small=10000;
			for (int v=1;v<n;v++)
			{
				int a=0;
				j=0;
				int big=0;
				while (j<k&&a<60)
				{
					if (j==-1||s[v][a]==cur[j])
					{
						j++;
						a++;
						big=max(big,j);
					}
					else
						j=next[j];
				}

				//cout<<big<<' ';
				small=min(small,big);
				//cout<<"  "<<small<<' ';

			}
			//cout<<endl;
			
			if (small>ans)
			{
				ans=small;
				for (j=0;j<small;j++)
					c[j]=cur[j];
				c[small]='\0';
			}
			if (small==ans)
			{
				char temp[66];
				for (j=0;j<small;j++)
					temp[j]=cur[j];
				temp[small]='\0';
				if (strcmp(temp,c)<0)
					strcpy(c,temp);
			}
			
		}
//		cout<<ans<<endl;

		if (ans<3)
			cout<<"no significant commonalities"<<endl;
		else
			cout<<c<<endl;
	}

	return 0;
}

Blue Jeans

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7922		Accepted: 3309

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