POJ 3080 Blue Jeans

Blue Jeans

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3080
64-bit integer IO format: %lld      Java class name: Main

 

 

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

 

解题：暴力查找长度最长，字典序最小的公共子串。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
char str[12][100];
int n,fail[100],len = 60;
void getFail(char *s,int m) {
    fail[0] = fail[1] = 0;
    for(int i = 1; i < m; i++) {
        int j = fail[i];
        while(j && s[i] != s[j]) j = fail[j];
        fail[i+1] = str[i] == str[j]?j+1:0;
    }
}
bool kmp(char *s,int m,int k) {
    getFail(s,m);
    for(int i = 0,j = 0; i < len; i++) {
        while(j && s[j] != str[k][i]) j = fail[j];
        if(str[k][i] == s[j]) {
            j++;
            if(j == m) return true;
        }
    }
    return false;
}
int main() {
    char tmp[100],ans[100];
    int i,j,k,t;
    bool flag;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        flag = false;
        for(i = 0; i < n; i++)
            scanf("%s",str[i]);
        len = strlen(str[0]);
        strcpy(ans,str[0]);
        for(i = len; i > 2 && !flag; i--) {
            for(j = 0; j+i <= len; j++) {
                strncpy(tmp,str[0]+j,i);
                tmp[i] = '';
                for(k = 1; k < n; k++)
                    if(!kmp(tmp,i,k)) break;
                if(k == n) {
                    flag = true;
                    if(strcmp(ans,tmp) > 0) strcpy(ans,tmp);
                    else strcpy(ans,tmp);
                }
            }
        }
        flag?puts(ans):puts("no significant commonalities");
    }
    return 0;
}