E - Blue Jeans
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
题目看起来还挺恐怖的，求最长公共子序列。第一想法是多匹配，AC自动机？感觉可以做。但是仔细一看数据，emmm这不是直接暴力就好了嘛，那哈希优化一下字符串的比较，就。好了。值得一提的是，当长度相同的时候要求输出字典序更小的。所以还是用string保存了一下答案
AC代码：
#include<algorithm>
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<vector>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#define endl '\n'
#define sc(x) scanf("%d",&x)
using namespace std;
typedef unsigned long long ULL;
const int size=65;
ULL H[10][size],P[size];
int Val[256];
char s[10][size];
void Hash(int len,int rank)
{
int LA=23333;
H[rank][0]=0;
P[0]=1;
for(int i=1;i<=len;i++)
{
H[rank][i]=H[rank][i-1]*LA+Val[s[rank][i]];
P[i]=P[i-1]*LA;
}
}
ULL code(int l,int r,int rank)
{
return H[rank][r]-H[rank][l-1]*P[r-l+1];
}
void init()
{
Val['A']=1;
Val['T']=2;
Val['G']=3;
Val['C']=4;
}
int main()
{
init();
int n;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(H,0,sizeof(H));
int i;
for(i=0;i<n;i++)
{
scanf("%s",s[i]+1);
Hash(60,i);
}
int maxl,maxr;
string ans="",ori(s[0]+1);
for(int l=1;l<=60;l++)
{
for(int i=1;i+l-1<=60;i++)
{
ULL sta=code(i,i+l-1,0);
int flag=0;
for(int k=1;k<n;k++)
{
flag=0;
for(int j=1;j+l-1<=60;j++)
{
ULL Sta=code(j,j+l-1,k);
if(Sta==sta) flag=1;
}
if(!flag)break;
}
if(flag)
{
string temp=ori.substr(i-1,l);
if(temp.size()>ans.size()) ans=temp;
else if(temp.size()==ans.size()&&temp<ans) ans=temp;
// break;
}
}
}
if(ans.length()<3) printf("no significant commonalities");
else cout<<ans;
cout<<endl;
}
return 0;
}