zoukankan      html  css  js  c++  java
  • POJ 3080 Blue Jeans

    Blue Jeans
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 9644   Accepted: 4064

    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
    题目大意:在一组60个字母组成的字符串中寻找它们的最长公共子序列,如果最长的有多个则打印其中按字典序排列最小的序列。
    #include <stdio.h>
    #include <iostream>
    #include <string.h>
    #include <stdlib.h>
    using namespace std;
    
    void init(char *pstr)
    {
        for (int i = 0; i < 64; i++)
        {
            pstr[i] = 'z';
        }
        pstr[64] = '';
    }
    
    int main()
    {
        char str[11][65];
        char strtemp[65];
        char strans[65];
        int N;
        int ncase;
        int flag;
        bool bfind = false;
        scanf("%d", &ncase);
        int nLen = 0;
        for (int i = 0; i < ncase; i++)
        {
            init(strans);
            flag = 0;
            bfind = false;
            scanf("%d", &N);
            for (int j = 0; j < N; j++)
            {
                scanf("%s", str[j]);
            }
            nLen = strlen(str[0]);
            for (int j = nLen; j >= 3; j--)
            {
                if (bfind)
                {
                    break;
                }
                for (int k = 0; k + j <= nLen; k++)
                {
                    strncpy(strtemp, str[0] + k, j);
                    strtemp[j] = '';
                    for (int l = 1; l < N; l++)
                    {
                        if (0 == strstr(str[l], strtemp))
                        {
                            break;
                        }
                        if (l == N - 1)
                        {
                            flag = 1;
                            if (strcmp(strtemp, strans) < 0)
                            {
                                strcpy(strans, strtemp);
                            }
                        }
                    }
                    if (flag && k + j == nLen)
                    {
                        printf("%s
    ", strans);
                        bfind = true;
                        break;
                    }
                }
            }
            if (!flag)
            {
                printf("no significant commonalities
    ");
            }
        }
        return 0;
    }
  • 相关阅读:
    Navicat15 for Mysql激活教程
    Overview
    NoSQL之一:Memcached
    Git学习(二):Git的初步使用
    Git学习(一):版本控制介绍及安装
    Docker学习(一):容器介绍
    ElasticStack学习(十):深入ElasticSearch搜索之QueryFiltering、多/单字符串的多字段查询
    ElasticStack学习(九):深入ElasticSearch搜索之词项、全文本、结构化搜索及相关性算分
    ElasticStack学习(八):ElasticSearch索引模板与聚合分析初探
    ElasticStack学习(七):ElasticSearch之Mapping初探
  • 原文地址:https://www.cnblogs.com/lzmfywz/p/3161767.html
Copyright © 2011-2022 走看看