poj 3080 Blue Jeans

最新推荐文章于 2020-06-16 18:36:58 发布

转载最新推荐文章于 2020-06-16 18:36:58 发布 · 65 阅读

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原文链接：http://www.cnblogs.com/GadyPu/p/4773468.html

本文介绍了一种用于比对多个DNA序列，寻找最长公共子序列的算法。该算法通过暴力枚举的方式，针对每一对DNA序列找出最长的共同序列，特别适用于基因研究中的序列比对任务。

题目连接

http://poj.org/problem?id=3080

Blue Jeans

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

字符串暴力枚举。。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
using std::map;
using std::min;
using std::find;
using std::pair;
using std::vector;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 61;
typedef char Word[61];
Word str1, str2, st[11];
void solve(int n) {
    bool f;
    int i, j, k, len = 0;
    for (i = 0; i < N - 3; i++) {
        for (j = 3; j < N - i; j++) {
            strncpy(str1, st[0] + i, j); str1[j] = '\0';
            f = true;
            for (k = 1; k < n; k++) {
                if (!strstr(st[k], str1)) { f = false; break; }
            }
            if (f && j > len) {
                len = j;
                strcpy(str2, str1);
            }
            if (f && j == len) {
                if (strcmp(str1, str2) < 0) strcpy(str2, str1);
            }
        }
    }
    puts(!len ? "no significant commonalities" : str2);
}
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w+", stdout);
#endif
    int t, n;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        rep(i, n) scanf("%s", st[i]);
        solve(n);
    }
    return 0;
}

转载于:https://www.cnblogs.com/GadyPu/p/4773468.html