format oci ont can for while fixed dataset man

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
題意：找最長的公共字串，長度相同就找最小的（這一點wa了我13遍！！！）
題解：kmp或者直接暴力列舉
kmp：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-9;
const int N=10+5,maxn=60+5,inf=0x3f3f3f3f;

string s[N];
int Next[maxn];

void getnext(string str,int slen)
{
    int k=-1;
    Next[0]=-1;
    for(int i=1;i<slen;i++)
    {
        while(k>-1&&str[k+1]!=str[i])k=Next[k];
        if(str[k+1]==str[i])k++;
        Next[i]=k;
    }
}
bool kmp(string ptr,int plen,string str,int slen)
{
    int k=-1;
    for(int i=0;i<plen;i++)
    {
        while(k>-1&&str[k+1]!=ptr[i])k=Next[k];
        if(str[k+1]==ptr[i])k++;
        if(k==slen-1)return 1;
    }
    return 0;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
 //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    int t,n;
    cin>>t;
    while(t--){
        cin>>n;
        for(int i=0;i<n;i++)cin>>s[i];
        string ans="";
        for(int i=1;i<=s[0].size();i++)//長度
        {
            for(int j=0;j<=s[0].size()-i;j++)//起點
            {
                string op=s[0].substr(j,i);
                getnext(op,op.size());
                bool flag=0;
                for(int k=1;k<n;k++)
                    if(!kmp(s[k],s[k].size(),op,op.size()))
                        flag=1;
                if(!flag)
                {
                    if(ans.size()<op.size())ans=op;
                    else if(ans.size()==op.size())ans=min(ans,op);
                }
            }
        }
        if(ans.size()<3)cout<<"no significant commonalities"<<endl;
        else cout<<ans<<endl;
    }
    return 0;
}

View Code

暴力：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-9;
const int N=10+5,maxn=60+5,inf=0x3f3f3f3f;

int Next[maxn];

void getnext(string str,int slen)
{
    int k=-1;
    Next[0]=-1;
    for(int i=1;i<slen;i++)
    {
        while(k>-1&&str[k+1]!=str[i])k=Next[k];
        if(str[k+1]==str[i])k++;
        Next[i]=k;
    }
}
bool kmp(string ptr,int plen,string str,int slen)
{
    int k=-1;
    for(int i=0;i<plen;i++)
    {
        while(k>-1&&str[k+1]!=ptr[i])k=Next[k];
        if(str[k+1]==ptr[i])k++;
        if(k==slen-1)return 1;
    }
    return 0;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
 //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    int t,n;
    cin>>t;
    while(t--){
        cin>>n;
        string str[15];
        for(int i=0;i<n;i++)cin>>str[i];
        string res="";
        for(int i=3;i<=60;i++)
        {
            for(int j=0;j<=60-i;j++)
            {
                string tem=str[0].substr(j,i);
             //   getnext(op,op.size());
                bool flag=1;
                for(int k=1;k<n;k++)
                    if(str[k].find(tem)==string::npos)
                    {
                        flag=0;
                        break;
                    }
                if(flag&&res.size()<tem.size())res=tem;
                else if(flag&&res.size()==tem.size()&&res>tem)res=tem;
            }
        }
        if(res=="")cout<<"no significant commonalities"<<endl;
        else cout<<res<<endl;
    }
    return 0;
}

View Code

poj3080kmp或者暴力