poj 3080 Blue Jeans (暴力枚举子串+kmp)

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

题意:给你m组DNA 要求你找到 最长公共的子串

思路：以第一个字符串为准 枚举 起点为 j 长度为 i 的子串 然后对其他字符串进行匹配（这个算法也是比较慢了）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
string s[10];
int dp[107];
void getnext(string t){
    dp[1]=0;
    for(int i=2,j=0;i<=t.length();i++){
        while(j>0&&t[i-1]!=t[j]) j=dp[j];
        if(t[i-1]==t[j]) j++;
        dp[i]=j;
    }
}
bool kmp(string t,string p){
    int len1=p.length();
    int len2=t.length();
    for(int i=1,j=0;i<=len1;i++){
        while(j>0&&p[i-1]!=t[j]) j=dp[j];
        if(p[i-1]==t[j]) j++;
        //cout<<j<<endl;
        if(j==len2)
        return true;
    }
    return false;
}
int main(){
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    while(n--){
        int m;
        cin>>m;
        for(int i=0;i<m;i++)
            cin>>s[i];
        int len=s[0].length();
        string ans="";
        for(int i=1;i<=len;i++){
            for(int j=0;j<=len-i;j++){
                string temp=s[0].substr(j,i);
                getnext(temp);
                bool jug=1;
                for(int k=1;k<m;k++)
                    jug&=kmp(temp,s[k]); //一假必假 
                if(jug){
                    if(ans.size()<temp.size()) ans=temp;
                    else if(ans.size()==temp.size()) ans=min(ans,temp);
                }
            }
            //cout<<ans<<endl;
        }
        if(ans.size()<3) cout<<"no significant commonalities"<<endl;
        else cout<<ans<<endl;
    }
    return 0;
}