POJ 3080 Blue jeans

Blue Jeans

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15090		Accepted: 6696

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

題意就是求最多三個字符串的最長的公共子序列的問題。

考慮用後綴數組的思維求解，從第一個字符開始依次向後移動，取後綴數組， 然後拿到後兩個字符串中進行比較，取兩者匹配長度的最小值，最後取所有後綴數組中匹配長度的最大值作爲結果。

一共就只有最多三個字符串，且每個字符串都只有60個字符，所以這麼做是完全可以的。

代碼如下：

/*************************************************************************
	> File Name: Blue_Jeans.cpp
	> Author: Zhanghaoran
	> Mail: [email protected]
	> Created Time: Thu 26 Nov 2015 06:21:10 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

char str[10][150];
int nexti[150];
char cmp[150];
int len;
int T;
int n;
int ans;
void preKMP(){
    int i, j;
    i = 0;
    j = nexti[0] = -1;
    while(i < len){
        if(j != -1 && cmp[i] != cmp[j]){
            j = nexti[j];
        }
        else if(cmp[++ i] == cmp[++ j])
            nexti[i] = nexti[j];
        else 
            nexti[i] = j;
    }

    //for(int i = 0; i <= m; i ++){
    //    cout << kmpnext[i] << " ";
    //}
    //cout << endl;
}
void Kmp_Count(){
    preKMP();
    int i, j, res;
    ans = 110;
    for(int temp = 1; temp < n; temp ++){
        i = 0;
        j = 0;
        res = 0;
        while(i < 60 && j < len){
            if(j == -1 || str[temp][i] == cmp[j]){
                i ++;
                j ++;
            }
            else 
                j = nexti[j];
            if(j > res)
                res = j;
        }
        if(res < ans)
            ans = res;

    }
}

int main(void){
    scanf("%d", &T);
    char ss[150];
    while(T --){
        scanf("%d", &n);
        for(int i = 0; i < n; i ++)
            scanf("%s", str[i]);
        int res = 0;
        for(int i = 0; i <= 57; i ++){
            strcpy(cmp, str[0] + i);
            len = 60 - i;
            Kmp_Count();                    
            if(res < ans){
                res = ans;
                strncpy(ss, str[0] + i, res);
                ss[res] = '\0';
            }
            else if(ans == res){
                char tt[150];
                strncpy(tt, str[0] + i, res);
                tt[res] = '\0';
                if(strcmp(tt, ss) < 0)
                    strcpy(ss, tt);
            }
        }
        if(res >= 3)
            cout << ss << endl;
        else 
            cout << "no significant commonalities" << endl;
    }
    return 0;
}